Question

A projectile is projected from the origin with a velocity of 45.0 m/s at an angle of 30.0 degrees above the horizontal. What is the velocity when the projectile hits the ground?

Answer 1

Answer:

velocity = 44.98 m/s

Explanation:

given data

velocity = 45.0 m/s

angle = 30.0 degrees

solution

first we get here time that is

time = 2u × sin θ ÷ g   .............1

time = 2×45 × sin 30 ÷ 9.8

time = 4.59 sec

and

velocity = $\sqrt{vx^2+vy^2}$   .....................2

here

vx = 45 cos30 = 38.97 m/s

vy = uy - gt

vy = 45 sin30 - 9.8 ×  4.59  = -22.48 m/s

so velocity will be

velocity = $\sqrt{38.97^2+22.48^2}$

velocity = 44.98 m/s