Question

What speed must you toss a ball straight up so that is takes 4.5 S back to get back to you? How high would it go?

Answer 1

Answer on Question 62059, Physics, Mechanics, Relativity

Question:

What speed must you toss a ball straight up so that it takes $4.5\,s$ to get back to you? How high would it go?

Solution:

a) We can find the initial velocity of the ball from the kinematic equation:

$$v = v_0 + g t,$$

here, $v = 0\,ms^{-1}$ is the final velocity of the ball, $v_0$ is the initial velocity of the ball, $g = -9.8\,ms^{-2}$ is the acceleration due to gravity (we take the upwards to be the positive direction, therefore $g$ will be with sign minus), $t = 4.5\,s / 2 = 2.25\,s$ is the time that needs the ball to reach the maximum height ($2.25\,s$ to go up and $2.25\,s$ to come down).

Then, from this formula we can calculate $v_0$:

$$v_0 = -g t = -(-9.8\,ms^{-2}) \cdot 2.25\,s = 22.05\,ms^{-1}.$$

b) We can find how high would the ball go from the kinematic equation:

$$h = v_0 t + \frac{1}{2} g t^2 = 22.05\,ms^{-1} \cdot 2.25\,s + \frac{1}{2} \cdot (-9.8\,ms^{-2}) \cdot (2.25\,s)^2 = 24.8\,m.$$

Answer:

a) $v_0 = 22.05\,ms^{-1}$.

b) $h = 24.8\,m$.

https://www.AssignmentExpert.com