Question

A body is projected at an angle of 30degrees to the horizontal with a speed of 30m/s. What will be the angle with the horizontal after 1.5sec. Take g as 10m/s^2

Answer 1

Given Information:

Launch angle of projectile = 30°

Initial velocity = V₀ = 30 m/s

Acceleration due to gravity = g = 10 m/s²

Required Information:

Angle with the horizontal after 1.5 sec = ?

Answer:

The angle of the projectile to the horizontal after t = 1.5 seconds is 0°

Explanation:

The horizontal component of the velocity is given by

`Vx = V_0 \cos(\theta)`

Where V₀ is the initial velocity and θ is the launch angle

The vertical component of the velocity is given by

`Vy = V_0 \sin(\theta) - gt`

Where V₀ is the initial velocity, θ is the launch angle, g is the acceleration due to gravity and t is the time.

So after t = 1.5 sec

The horizontal component of the velocity is

`Vx = V_0 \cos(\theta) \\\\Vx = 30 \cos(\30) \\\\Vx = 30 * 0.866\\\\Vx = 25.981 \: m/s`

And the vertical component of the velocity is

`Vy = V_0 \sin(\theta) - gt \\\\Vy = 30 \sin(30) - 10 * 1.5 \\\\Vy = 30(0.5) - 10 * 1.5 \\\\Vy = 15 - 15 \\\\Vy = 0 \: m/s \\\\`

The angel is

`\tan(\theta) = (0)/(25.981) \\\\\theta= \tan^(-1)( (0)/(25.981)) \\\\\theta= 0`

Therefore, the angle of the projectile to the horizontal after t = 1.5 seconds is 0°