Question

A person throws a ball with an initial velocity of 121 m/s and the ball takes 4 s to reach its maximum height. What was the maximum height reached?

Answer 1

Ok, imagine that it is a magical ball that always want to move to west in 4 m/s^2. You throw that ball to east at 22 m/s. In 6 seconds, that magical ball will turn back to you. I will fracture the distances of that ball travelled by it's vector (east and west)

 (until the ball stop to return to you)

Final velocity = V - a.t

0 = 22 - 4t

t = 11/2 s

Distance = V.t - 1/2 a.t^2

= 22 . 11/2 - 1/2 . 4 . (11/2)^2

= 121 - 121/2

= 121/2 m

t = 6 - 11/2

= 1/2 s

Distance = V.t + 1/2 a.t^2

= 0 . 1/2 + 1/2 . 4 . (1/2)^2

= 1/2 m

Answer 2

here u= 121 m/s
t = 4 s
a =g ( acc due to gravity) = -9.8m/s²
at maximum height v= 0
from v= u + at
0= 121 + (- 9.8 )x t
= -121 = - 9.8 x t
-121 ÷ - 9.8 = t
t = you can find