Question

A body projected upward from the level ground at an angle of 50with the horizontal has an initial speed of 40 m/s. (a) How long will it take to hit the ground? (b) How far from the starting point will it strike?

Answer 1

Answer:

For the 1st part we can use:

v=u+at

If we consider only the vertical component of motion this becomes:

0=vsin50−gt

∴

t= 40 sin 50 / 9.8

=3.126 s

This is the time taken for the body to reach its maximum height.

To fall to the ground takes the same amount of time so we can say:

t tot =3.126×2

= 6.25s

To get the range we consider the horizontal component of motion.

The velocity in the x direction is constant so we can say:

s=d/t

∴

d=s×t

=40cos50×6.25

= 160.7m

hope it will help you.....