Question

Projectile Motion - Rocket Clears Wall - General Launch Angle

A rocket is fired at an initial speed v0 = 165.0 m/s from ground level, at an angle θ = 50° above the horizontal.

A wall is located at d = 67.0 m. Its heigh is h = 47.0 m. Ignore air resistance.

The magnitude of the gravitational acceleration is 9.8 m/s2.

Choose the RIGHT as positive x-direction. Choose UPWARD as psotitive y-direction

Find v0x, the x component of the initial velocity (in m/s) . Keep 2 decimal places.

Find v0y, the y component of the initial velocity (in m/s) . Keep 2 decimal places.

You will calculate the time at which the rocket flies over the wall.

To find this time, should you use the horizontal (x) motion or the vertial (y) motion?

(a) At what time (in seconds) (after being fired) does the rocket fly over the wall? Keep 4 decimal places.

Note: when the rocket flies over the wall, in general, the height of the rocket is not equal to the maximum height.

(b) What is the rocket's height (in meters) when it flies over the wall? Keep 3 decimal places.

(c) By how much (in meters) does the rocket clear the top of the wall? Keep 3 decimal places.

(d) What is the rocket's maximum height (in meters) above the ground? Keep 2 decimal places.

Answer 1

Given that Rocket is fired at V0 = 165.0 m/sec at 50 deg above the horizontal

V0x = Initial horizontal velocity = V0*cos = 165.0*cos 50 deg = 106.06 m/s

V0y = Initial vertical velocity = V0*sin = 165.0*sin 50 deg = 126.40 m/s

to find the time at which rocket flies over the wall, we use horizontal motion equation.

A.

Range in projectile motion is given by:

R = V0x*T

Since there is no acceleration in horizontal direction so it's horizontal velocity will remain constant

T = time when rocket is above the wall = ?

R = Range = distance of wall from launching point = 67.0 m

T = R/V0x = 67.0/106.06

T = 0.6317 sec

Part B.

At this time rocket's vertical height will be given by 2nd kinematic equation in vertical direction:

h1 = V0y*t + (1/2)*ay*t^2

ay = acceleration in vertical direction = -g = -9.8 m/s^2 (-ve because g is downward)

h1 = 126.40*0.6317 + (1/2)*(-9.8)*0.6317^2

h1 = 77.892 m = Height of rocket from ground when it is above wall

Part C.

h = height of wall = 47.0 m

dh = h1 - h = 77.892 - 47.0 = 30.892 m

So rocket clears the top of wall by 30.892 m distance.

Part D.

Max height in projectile motion is given by:

H_max = V0y^2/(2*g)

H_max = 126.40^2/(2*9.8)

H_max = 815.15 m

Let me know if you've any query.