Question

A projectile returns to its original height 4.08 s after being launched, during which time it travels 76.2 m horizontally. If air resistance can be neglected, what was the projectile's initial speed?

Answer 1

Answer:27.35 m/s

Explanation:

Given

Time of Flight of Projectile T=4.08 s

Range of Projectile =76.2 m

Time Of Flight of Projectile is given by

$T=\frac{2u\sin \theta }{g}----------1$

where u=initial Velocity

$\theta =$Launch angle

g=acceleration due to gravity

Range is given by $R=\frac{u^2\sin 2\theta }{g}------2$

divide 1 and 2

$\frac{R}{T}=\frac{u^2\sin 2\theta }{g}\times \frac{g}{2u\sin \theta }$

$\frac{R}{T}=u\cos \theta ------3$

$u\sin \theta =\frac{Tg}{2}------4$

squaring and adding 3 & 4 we get

$u^2(\cos ^2\theta +\sin ^2\theta )=(\frac{R}{T})^2+(\frac{Tg}{2})^2$

$u^2=(19.992)^2+(18.676)^2$

$u=\sqrt{748.49}$

$u=27.35 m/s$