27.03.2020

g You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 37° from the edge of the building with an initial velocity of 21 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?

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Step-by-step answer

24.06.2023, solved by verified expert
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h = 21.5 m

Explanation:

First of all, we define a pair of coordinate axes along the horizontal and vertical direction, calling x-axis to the horizontal and y-axis to the vertical, with the origin in the point where the ball is kicked.Neglecting air resistance, the only influence on the ball once kicked is due to gravity, so the ball is accelerated by the Earth with a constant value of -9.8 m/s2 (assuming the upward direction as positive).So, we can use the kinematic equation for displacement for the vertical direction, as follows:

       g You are walking around your neighborhood and, №17886370, 27.03.2020 17:09

Since the ball is kicked at an angle of 37º from the edge of the building, at an initial velocity of 21 m/s, we can find the horizontal and vertical initial speeds as follows:

       g You are walking around your neighborhood and, №17886370, 27.03.2020 17:09

       g You are walking around your neighborhood and, №17886370, 27.03.2020 17:09

In the horizontal direction, since gravity has no component in this direction, the ball moves at a constant speed, equal to v₀ₓ.Applying the definition of average velocity, since we know the horizontal distance traveled, we can find the total time that the ball was in the air, as follows:

       g You are walking around your neighborhood and, №17886370, 27.03.2020 17:09

Replacing (4) and (3) in (1), we can find the total vertical displacement, which is equal to the height of the building, as follows:

     g You are walking around your neighborhood and, №17886370, 27.03.2020 17:09

⇒ h = -(-21.5m) = 21.5 m
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Physics
Step-by-step answer
P Answered by PhD

h = 21.5 m

Explanation:

First of all, we define a pair of coordinate axes along the horizontal and vertical direction, calling x-axis to the horizontal and y-axis to the vertical, with the origin in the point where the ball is kicked.Neglecting air resistance, the only influence on the ball once kicked is due to gravity, so the ball is accelerated by the Earth with a constant value of -9.8 m/s2 (assuming the upward direction as positive).So, we can use the kinematic equation for displacement for the vertical direction, as follows:

       \Delta y = v_{oy}* t -\frac{1}{2}*g*t^{2}   (1)

Since the ball is kicked at an angle of 37º from the edge of the building, at an initial velocity of 21 m/s, we can find the horizontal and vertical initial speeds as follows:

       v_{ox} = v* cos 37 = 21 m/s * cos 37 = 16.8 m/s (2)

       v_{oy} = v* sin 37 = 21 m/s * sin 37 = 12.6 m/s (3)

In the horizontal direction, since gravity has no component in this direction, the ball moves at a constant speed, equal to v₀ₓ.Applying the definition of average velocity, since we know the horizontal distance traveled, we can find the total time that the ball was in the air, as follows:

       t = \frac{\Delta x}{v_{ox} }  = \frac{63m}{16.8m/s} = 3.75 s (4)

Replacing (4) and (3) in (1), we can find the total vertical displacement, which is equal to the height of the building, as follows:

     -h = 12.6m/s* 3.75s -\frac{1}{2}*(9.8m/s2)*(3.75s)^{2} = -21.5 m (5)

⇒ h = -(-21.5m) = 21.5 m
Physics
Step-by-step answer
P Answered by PhD
Answer:
7.25 secs.

Explanation:
First find the distance it takes to stop
s = [v^2-u^2]/2a = 0^2 - 8.7^2/2[-2.4] = 8.7^2/4.8
Next find the time it takes to go that distance , s = ut +[1/2] at^2
8.7^2/4.8 = 8.7t +[1/2] [ -2.4]t^2 , rearrange and
t^2 -[8.7/1.2]+ 8.7^2/[(1.2)(4.8)]=0 complete the square
[t - (8.7/2.4)]^2=0
t = 8.7/2.4 = 3.625 secs
At this stage the deceleration will push the object back in the direction it came from for another 3.625 secs when it will be 8.7 m/s again
Total time , T =2t = 7.25 secs.

Note:
The term differential is used in calculus to refer to an infinitesimal (infinitely small) change in some varying quantity. For example, if x is a variable, then a change in the value of x is often denoted Δx (pronounced delta x). The differential dx represents an infinitely small change in the variable x.
Physics
Step-by-step answer
P Answered by PhD
First sum applied the Newton's second law motion: F = ma
Force = mass* acceleration
This motion define force as the product of mass times Acceleration (vs.Velocity). Since acceleration is the change in velocity divided by time,
force=(mass*velocity)/time
such that, (mass*velocity)/time=momentum/time
Therefore we get mass*velocity=momentum
Momentum=mass*velocity
Elephant mass=6300 kg; velocity=0.11 m/s
Momentum=6300*0.11
P=693 kg (m/s)
Dolphin mass=50 kg; velocity=10.4 m/s
Momentum=50*10.4
P=520 kg (m/s)
The elephant has more momentum(P) because it is large.
Physics
Step-by-step answer
P Answered by PhD
The change in temperature is 9.52°CExplanation:Since, the heat supplied by the electric kettle is totally used to increase the temperature of the water.Thus, from the law of conservation of energy can be stated as:Heat Supplied by Electric Kettle = Heat Absorbed by WaterHeat Supplied by Electric Kettle = m C ΔTwhere,Heat Supplied by Electric Kettle = 20,000 JMass of water = m = 0.5 kgSpecific Heat Capacity of Water = C = 4200 J/kg.°CChange in Temperature of Water = ΔTTherefore,20,000 J = (0.5 kg)(4200 J/kg.°C) ΔTΔT = 20,000 J/(2100 J/°C)ΔT = 9.52°C
Physics
Step-by-step answer
P Answered by PhD
Weight of barbell (m) = 100 kg
Uplifted to height (h) = 2m
Time taken= 1.5 s
Work done by Jordan = potential energy stored in barbell = mgh
= 100×2×9.8
= 1960J
Power = energy/time
= 1960/1.5
1306.67watts
Physics
Step-by-step answer
P Answered by PhD
Weight of jasmine (m) = 400 N
Height climbed on wall (h) = 5m
Total time taken in climbing = 5 sec
Work done in climbing the wall = rise in potential energy = mgh
= 400×9.8×51
= 19600J
Power generated by Jasmine = potential energy / time
= 19600/5
= 3920Watts
Physics
Step-by-step answer
P Answered by PhD
The horizontal and vertical motions of balloons are independent from each other.
Let vertical component of initial velocity U' horizontal component of initial velocity U"
Time of landing (t) is found with the help of vertical motion.
Since vertical component of initial velocity of balloon is zero(U' = 0)
From equation h = U't + 1/2gt^2
h = 1/2gt^2
t = √(2h/g)
t = √( 2×150/9.8)
t = 5.53 sec
Horizontal velocity = 50m/s
Horizontal range of balloon, R = U"t
= 50× 5.53
= 27.65m
So the balloon will go 27.65 metre away from the bridge
Physics
Step-by-step answer
P Answered by PhD
Gravity acceleration (g) = 9.8m/s^2
Time (t) = 3sec
Acceleration = velocity/time
Velocity = acceleration×time
= 9.8×3
= 29.4m/s
Physics
Step-by-step answer
P Answered by PhD
Initial velocity (u) = 0
Time taken = 4.5 seconds
Gravitational acceleration (g) = 9.8m/s^2
By the second equation of motion under gravity,
The distance that object fell down (h)
h = ut + (1/2)gt^2
h = 0×4.5 + (1/2)×9.8×(4.5)^2
h = 99.225 m

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