Question

A projectile is projected with velocity of 50m/s at an angle of 60^° with the horizontal from the ground. The time after which is velocity will make an angle of 45^° with the horizontal is

1)2.5s

2)1.83s

3)2.37s

4)3.72s

Answer 1

To determine the angle of the velocity, we use the following equation.
Tan θ = Vertical velocity ÷ Horizontal velocity

As the particle moves from its initial position to its final position, its vertical velocity decreases at the rate of 9.8 m/s each second. During this time its horizontal velocity is constant. To determine the time when the angle is 45˚, we need to determine the object’s vertical velocity at this angle. Let’s determine the vertical and horizontal components of its initial velocity.

Vertical = 50 × sin 60˚, Horizontal = 50 × cos 60˚ = 25 m/s

Since horizontal velocity is constant, let’s put 25 m/s and 45˚ into the tangent equation.

Tan 45 = Vertical velocity ÷ 25
Vertical velocity = 25 m/s
To determine the time for the vertical velocity to decrease from its initial value to 25 m/s, use the following equation.

vf = vi – a × t
a = 9.8
25 = 50 × sin 60˚ – 9.8 × t
t = (25 – 50 × sin 60˚) ÷ -9.8
This is approximately 1.83 seconds.