Question

a baseball is thrown vertically upwards and returns to the thrower three seconds later find speed at which the ball us thrown​

Answer 1

Answer:

14.7 m/s

Explanation:

Since the motion of the ball is a uniformly accelerated motion, we can find the initial speed by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the vertical displacement

u is the initial velocity

t is the time

a is the acceleration

In this problem, we have:

s = 0 (because at the end of the motion, the ball returns to its original position)

u = ?

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

$t = 3 s$ is the total time of flight

Solving for u, we find the initial velocity of the ball:

$0=ut+\frac{1}{2}at^2\\u=-\frac{1}{2}at=-\frac{1}{2}(-9.8)(3)=14.7 m/s$