Question

A body is projected at angle 30° to horizontal with a velocity 50 m/s. It's time of flight is
g=10m/s2​

Answer 1

The time of flight is given by

T=

g

2usinθ

=

10

2×30×1/2

=3 s

Thus, after 1.5 s the body will be at the highest point.

So the direction of motion will be horizontal after 1.5 s, the angle with the horizontal is 0

∘

.

Answer 2

Answer:

5 sec

Explanation:

Given,

angle theta =30 degrees

initial velocity u=50 m/s

g=10m/s

we know that,

time of flight=2usintheta/g

=2×50×sin30/10

=100×1/2/10

=5sec