Question

Rodney throws a ball upward into the air with an initial velocity of 18 m/s. How high does it go and how long the ball is in the air before it comes back to his hand?

Answer 1

Given:

The initial velocity with which the ball is thrown in upward direction, u = 18 m/s

Take, g = 9.81 m/s2

We know that the final velocity of the ball at the highest point, v = 0

Let h be the height up to which it go.

Using the kinematic equation, v2 = u2 + 2as

Here, acceleration, a = -g and displacement, s = h

0 = u2 - 2gh

2gh = u2

h = u2 /2g= 182 /2 x 9.81 = 16.51 m

using the equation, v =u +at

we get, 0 = u – gt where t is time taken to reach the highest point

gt = u

t = u/g = 18 / 9.81 = 1.83 s

total time taken by the ball to return to the hand is T = 2t = 2 x 1.83 = 3.66 s

Answer: height, h = 16.51 m

                Time, T = 3.66 s

answered by: anonymous