Problem 2/77 Page 49, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A projectile is released at a height H above the ground with a velocity of magnitude v0 and making an angle of q0 above the horizontal.

Find: The height of the particle after it has traveled a horizontal distance L.

Relevant Parameters:

L = 200 ft
v0 = 140 ft/s and 120 ft/s
q0 = 8 degrees and 12 degrees
H = 7.5 ft
m = Projectile mass

1. Mechanical System = Projectile in flight

2. Free Body Diagram: Incomplete - would show only the projectile and the gravitational force.

3. Equations
a. S F = - m g j
b. m a = m ( ax i + ay j ) = m ( dvx/dt i + dvy/dt j )

4. Solve
dvx/dt = 0
dvy/dt = -g

Separating variables:
dvx = 0 dt
dvy = -g dt

Integrating both sides:

vx = Cx
vy = -g t + Cy

Using the facts that at t=0, vx = v0 cos(q0) and vy = v0 sin(q0), enables the evaluations of Cx and Cy.

vx = v0 cos(q0)
vy = -g t + v0 sin(q0)

Recalling the definition of velocity:

dx/dt = vx = v0 cos(q0)
dy/dt = vy = -g t + v0 sin(q0)

Separating variables:

dx = v0 cos(q0) dt
dy = { -g t + v0 sin(q0) } dt

Integrating both sides:

x = v0 cos(q0) t + Dx
y = -1/2 g t² + v0 sin(q0) t + Dy

Using the facts that at t=0, x=0 and y=H, enables the evaluations of Dx and Dy.

x = v0 cos(q0) t
y = -1/2 g t² + v0 sin(q0) t + H

To determine the time required to reach the desired x coordinate:

L = v0 cos(q0) td

td = L / v0 cos(q0) = 1.443 s and 1.704 s

To determine the height:

h = -1/2 g td² + v0 sin(q0) td + H =  2.10 ft and 3.27 ft