Question

A ball is thrown vertically upwards and rises to a height of 126.5 m. Calculate the

velocity by which the ball was thrown upwards and
(ii) the time taken by the ball to reach the highest point.
(Given isg=9.8ms-2)
​

Answer 1

Answer:

Given the height at which ball is thrown, h = 126.5 m.

1) To calculate the initial velocity of ball by which ball was thrown, use equation of motion,

v^{2} =u^{2} -2ghv2=u2−2gh   (upwards thrown)

Here, u is the initial velocity v is final velocity and h is height at which ball reached.

At height 126.5 m (maximum height) , the final velocity is zero.

Substitute the given values, we get

0 = u^{2} -2\times9.8 m/s^2\times 126.5 m0=u2−2×9.8m/s2×126.5m

u=\sqrt{2\times9.8 m/s^2\times 126.5 m}u=2×9.8m/s2×126.5m

u = 49.79 m/s

Thus, the velocity by which the ball was thrown is 49.79 m/s.

2) To calculate the time taken by the ball to reach the highest point, use equation of motion

v = u - gt

Substitute the values, we get

0 = 49.79-9.8m/s^2\times t0=49.79−9.8m/s2×t

t=\frac{49.79m/s}{9.8m/s^2}t=9.8m/s249.79m/s

t = 5.08 s.

Thus, the time taken by the ball to reach the highest point is 5.08 s.

Answer 2

Answer:

.............hope this helps.......