Question

Suppose that the gravitational acceleration on a certain planet is only
7.3 m/s². A space explorer standing on this planet throws a ball straight upward with an initial velocity of 11.2 m/s.
(a) What is the velocity of the ball 4.9 seconds after it is thrown?
To indicate direction, remember up is positive and down is negative.
(b) How much time elapses before the ball reaches the high point of its flight?
(c) What is the highest point of the ball from the ground?

Answer 1

Basically a projectile motion has two components namely the horizontal and vertical motion. In this problem, we will just focus on the vertical motion because the ball just traveled vertically.

Here the given values are

acceleration due to gravity = a = 7.3 m/s2

Initial velocity = u = 11.2 m/s

a) t = 4.9 s

Let the velocity of the ball after 4.9 s is v.

Using the equation of motion,

v = u + a t

v = 11.2 - 7.3 x 4.9

v = -24.57 m/s (means the ball falling downward at this time)

b) At the highest point, v = 0

Then

0 = 11.2 - 7.3 t

t = 11.2/7.3 = 1.534 s

c) Using v2 - u2 = 2aS

0 - 11.22 = -2*7.3*S

S = 8.6 m