Question

a ball is thrown upward with a speed of 39.2m/s calculate the maximum height it reaches and the time taken in reaching the maximum height

Answer 1

sorry to say you that I am not know your questions answered sorry

Answer 2

Given

u = 39.2 m/s

v = 0 m/s

g = -9.8m/s

We know

(i) $v^{2} - u^{2}$ = 2gh

0 - 1536.64 = 2 × (-9.8) × h

h = $\frac{-1536.64}{2 × (-9.8)}$

∴ h = 78.4 m

(ii) t = $\frac{v-u}{g}$

t = $\frac{0-39.2}{-9.8}$

t = $\frac{-39.2}{-9.8}$

∴ t = 4 s

Hope this helps :)