Physics, asked by borabeauty20, 1 day ago

A ball is thrown upward with a speed of 100m/s. Show that the ball hits the ground with the same speed after returning from highest point. Take g=10m/s^2, Neglect the air resistance.​

Answers

Answered by souravrajak2020
0

Explanation:

First, find the distance that the ball will reach in the air, this is the height

From

V^2 - Vo^2 = 2aS _____ Eq. 1

We solve for S:

S = (V^2 - Vo^2)/ 2a

At the maximum height V = 0, after substituting on the equation 1:

S = (- Vo^2) / 2a

Since a = -9.8 m/s^2 (negative upwards) and Vo = 100 m/s

S = (- (100 m/s)^2) / 2 (-9.8m/s^2) = 510.20 m

Hence, the distance that the object will fall is the same 510.20 m, when the object returns (as in free falling) Vo = 0 and g = 9.8 m/s^2 (positive downwards)

After solving the Equation 1 for V and the substitution of the former values

V = Root (Vo^2 + 2aS) ___ This formula is known as the gravitational speed!

V = Root( 0 + 2(9.8m/s^2)(510.20 m) ) = 99.99 m/s

Finally we round the result as V = 100 m/s

Good luck!

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