Question

A 0.90-kg ball is thrown with a speed of 9.0 m/s at an upward angle of 27 ∘.A) what is its speed at its highest point, andb)how high does it go?

Answer 1

Answer:a)8.01 m/s

Explanation:

Given

mass of ball $m=0.9 kg$

initial speed $u=9 m/s$

launch angle $\theta =27$

As the projectile reaches its highest point its vertical velocity component becomes zero and there will only be horizontal component

velocity at highest Point $v=u\cos \theta$

$v=9\cos 27$

$v=8.01 m/s$

(b)maximum height h

Maximum height is given by

$H_{max}=\frac{u^2\sin^2 \theta }{2g}$

$H_{max}=\frac{9^2\sin^2 (27)}{2\times 9.8}$

$H_{max}=0.85 m$