A ball is gently dropped from a height of 30m. If its velocity increase uniform at the rate of 10m/s with what velocity will it strikes the ground ? After what time will it strike the ground?

Explanation:

Need to FinD :The velocity of the ball with strikes the ground.The time of the ball after which it strikes the ground.

We know that,

The ball is dropped from the height of 30 m as stated in the question. Therefore, it will have a initial velocity of 0 m/s. The distance travelled by the ball will be 30 metres as the distance travelled by the ball is equal to the height of the tower and that is 30 metres.

So, by using the third equation of motion, we will find out the final velocity of the ball.

Therefore,

∴ Hence, the required final velocity of the ball with which it strikes the ground is 24.5 m/s. Since, we know that the initial velocity of the ball is 0 m/s and the acceleration of the ball is 10 m/s². So, by using the first equation of motion, we will find out the time of the ball after which it stikes the ground.

Therefore,

∴ Hence, the required time of the ball after which it strikes the ground is 2.45 seconds.

Options:
a. 0.08 meters
b. 0.16 meters
c. 0.32 meters
d. 1.8 meters

Answer:
b. 0.16 meters

Explanation:
In the picture

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.

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

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

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

Gravity acceleration (g) = 9.8m/s^2
Time (t) = 3sec
Acceleration = velocity/time
Velocity = acceleration×time
= 9.8×3
= 29.4m/s

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