A ball thrown vertically is caught by the thrower after 5.1s. Find the maximum height the ball reaches.

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

the answer is b i think

it is a negative acceleration