Question

A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 24.0 m/s from a height of 3.0 m. (a) How high does the ball rise from its original position? 29.38 Correct: Your answer is correct. m (b) How long does it take to reach its highest point? 2.44 Correct: Your answer is correct. s (c) How long does the ball take to hit the ground after it reaches its highest point? 2.44 Incorrect: Your answer is incorrect. s (d) What is the ball's velocity when it returns to the level from which it started?

Answer 1

Answer:

a) 29.36 m

b) 2.44 s

c) 2.57 s

d) 25.117 m/s

Explanation:

t = Time taken

u = Initial velocity = 24 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

b)

$v=u+at\\\Rightarrow 0=24-9.81\times t\\\Rightarrow \frac{-24}{-9.81}=t\\\Rightarrow t=2.44 \s$

Time taken by the ball to reach the highest point is 2.44 seconds

a)

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times 2.44+\frac{1}{2}\times -9.81\times 2.44^2\\\Rightarrow s=29.35\ m$

The highest point reached by the ball above its release point is 29.36 m

c) Total height is 3+29.35 = 32.35 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 32.35=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.35\times 2}{9.81}}\\\Rightarrow t=2.57\ s$

The ball reaches the ground 2.57 seconds after reaching the highest point

d)

$v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2117\ m/s$

The ball will hit the ground at 25.2117 m/s