Question

question 15

a person throws a ball upwards. it takes 10s for it to reach the high point and re - reach equilibrium point.

a) how long until it reaches the high point?

b) what is velocity at high point?

c) what is the initial velocity?

d) what is velocity at t = 2s?

Explanation:

Step1: Time to reach high - point

Since the time to go up and come back down (to the equilibrium point) is \(t = 10s\), and the time to go up is equal to the time to come down. Let \(t_{up}\) be the time to reach the high - point. Using the symmetry of vertical motion \(t = t_{up}+t_{down}\) and \(t_{up}=t_{down}\). So \(t_{up}=\frac{t}{2}\).
\(t_{up}=\frac{10s}{2}=5s\)

Step2: Velocity at high - point

At the highest point of vertical motion, the vertical component of velocity is \(0\). This is because the ball stops momentarily before starting to fall back down. So \(v = 0m/s\)

Step3: Initial velocity

Use the equation \(v = v_0 - gt\) (where \(v = 0\) at the highest point, \(t=t_{up}=5s\) and \(g = 9.8m/s^{2}\)). Rearranging for \(v_0\): \(v_0=v + gt\). Substituting \(v = 0\), \(t = 5s\) and \(g=9.8m/s^{2}\)
\(v_0=0+9.8\times5=49m/s\)

Step4: Velocity at \(t = 2s\)

Use the equation \(v=v_0 - gt\). Here \(v_0 = 49m/s\), \(t = 2s\) and \(g = 9.8m/s^{2}\)
\(v=49-9.8\times2=49 - 19.6=29.4m/s\)

Answer:

a) \(5s\)
b) \(0m/s\)
c) \(49m/s\)
d) \(29.4m/s\)