Question

A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It strikes a target 1.7 seconds later. What is the vertical distance from where the projectile was launched to where it hit the target.

Answer 1

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± $\frac{1}{2} at^2$               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± $\frac{1}{2} at^2$               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

From the question;

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - $\frac{1}{2} at^2$

h = 26.54(1.7) - $\frac{1}{2} (10)(1.7)^2$

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m