A ball thrown vertically is caught by the thrower after 5.1s. Find the maximum height the ball reaches. July 17, 2021 by Maris A ball thrown vertically is caught by the thrower after 5.1s. Find the maximum height the ball reaches.

Answer: h = 31.9 m Explanation: Since, the ball took 5.1 s in the air. Hence, the time to reach maximum height will be equal to the half of this value: t = 5.1 s /2 = 2.55 s Now, we use 1st equation of motion between the time of throwing and the time of reaching maximum height: Vf = Vi + gt where, Vf = Final Velocity = 0 m/s (since, ball momentarily stops at highest point) Vi = Initial Velocity = ? g = – 9.8 m/s² (negative sign for upward motion) Therefore, 0 m/s = Vi + (-9.8 m/s²)(2.55 s) Vi = 24.99 m/s Now, we use second equation of motion for height (h): h = Vi t + (0.5)gt² h = (24.99 m/s)(2.55 s) + (0.5)(-9.8 m/s²)(2.55 s)² h = 63.7 m – 31.8 m h = 31.9 m Reply

Answer:h = 31.9 m

Explanation:Since, the ball took 5.1 s in the air. Hence, the time to reach maximum height will be equal to the half of this value:

t = 5.1 s /2 = 2.55 sNow, we use 1st equation of motion between the time of throwing and the time of reaching maximum height:

Vf = Vi + gtwhere,

Vf = Final Velocity = 0 m/s (since, ball momentarily stops at highest point)

Vi = Initial Velocity = ?

g = – 9.8 m/s² (negative sign for upward motion)

Therefore,

0 m/s = Vi + (-9.8 m/s²)(2.55 s)Vi = 24.99 m/sNow, we use second equation of motion for height (h):

h = Vi t + (0.5)gt²h = (24.99 m/s)(2.55 s) + (0.5)(-9.8 m/s²)(2.55 s)²h = 63.7 m – 31.8 mh = 31.9 m