if a ball is thrown straight upward and is caught 5s after being thrown,how fast must it have been going when it left the person s hand?help

Hint: It takes the same amount of time to go up, as it does down, so use 2.5 as t. And your only acceleration will be gravity acting downwards.

Figure out a, and vi, then put them into one of your kinetic formulas and you should get your answer.

A ball is thrown vertically into the air and takes 5s to return to the thrower, from time of projection. What is the time it takes to reach maximum height ?

To determine the initial velocity of the ball when it left the person's hand, we can use the kinematic equation for vertical motion:

v = u + at

where:
v is the final velocity (0 m/s when the ball reaches its maximum height),
u is the initial velocity (what we're trying to find),
a is the acceleration due to gravity (-9.8 m/s^2 for objects near the surface of the Earth),
t is the time taken (5 seconds in this case).

Since the final velocity is 0 m/s when the ball reaches its maximum height, we can rewrite the equation as:

0 = u + (-9.8 m/s^2)(5s)

Rearranging the equation, we can solve for u:

u = -(-9.8 m/s^2)(5s)
u = 49 m/s

Therefore, the ball must have been thrown with an initial velocity of 49 m/s when it left the person's hand to be caught 5 seconds later.