Question

a projectile is fired with an initial velocity of 60 m/s upward at an angle of 30 degree to the horizontal from a point 80m above a level plain. what horizontal distance will it cover before it strikes the plain?

Answer 1

Answer

This is vertical speed component

u=60sin 30°=30m/s

Height h=80m

Using second equation of motion

S=ut+0.5at²

Putting all values

-80=30t-4.9t²

So this quadratic equation in t

So it's solutions

t² -6.12t-16.32=0

This is quadratic equation in t it's solutions are

$t=\frac{6.12±\sqrt{37.45+65.28}}{2}\\=\frac{6.12±\sqrt{102.73}}{2}\\=\frac{6.12±10.14}{2}$

By using positive sign then time

t=10. 26sec

So horizontal distance

D=60 cos 30° *10.26=533.12m