Question

17.A ball is thrown upwards with a speed of 39.2 m/s. Calculate the maximum height it reaches,

Answer 1

since the final velocity at topmost point is 0, maximum height reached is u^2/2g
u being the initial velocity
this max height reached is (39.2^2)/(2×9.8)=78.4m

Answer 2

When ball is thrown upward at highest point velocity of ball will be zero.
use, formula,
V² = u² + 2aS
Here,
V = 0 { final velocity }
u = 39.2 m/s
a = - g = -9.8 m/s²
So, 0 = (39.2)² +2(-9.8)S
$S = \frac{39.2^2}{19.6}$
$S = \frac{39.2*39.2}{19.6}$
S = 78.4 m
Hence, maximum height = 78.4 m