Question

When a ball is launched from the ground at a 45° angle to the horizontal, it falls back to the ground 50 m from the launch point. If it is launched at the same speed directly upward, (a) how high does it get? (b) How much longer does it spend in the air?

Answer 1

Answer:

Explanation:

Given

angle through which ball is launched$=45^{\circ}$

Range of ball=50 m

Range of projectile is $=\frac{u^2sin2\theta }{g}$

$50=\frac{u^2sin90}{9.8}$

u=22.136 m/s

If ball is thrown straight upward

$v^2-u^2=2as$

$0-(22.136)^2=2(-9.8)s$

$s=\frac{22.136^2}{2\times 9.8}$

s=25 m

(b)For Projectile time of flight is

$t=\frac{2usin\theta }{g}$

$t=\frac{2\times 22.136\times sin45}{9.8}$

t=3.19 s