A rocket is fired at a speed of 78.6 m/s from ground level, at an angle of 59.3° above the horizontal. The ro?
A rocket is fired at a speed of 78.6 m/s from ground level, at an angle of 59.3° above the horizontal. The rocket is fired toward an 12.8 m high wall, which is located 26.0 m away. By how much does the rocket clear the top of the wall?
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here's a quick way to solve
the usual drill is to divide the motion into horizontal (X) motion
and vertical (Y) motion
and these equations
Vx= Vi cosθ
Vy= Vi sinθ
Vx= X/t
Y= Vy*t + 1/2 g t^2
yeah, it gets tedious
so
I combined the equations into one that makes quicker work of most trajectory problems
∆Y = X*tanθ + ( (g*X^2) / 2(Vi*cosθ)^2 )
if the projectile comes back to ground level, ∆Y = 0 which simplifies things a lot
otherwise higher is positive and lower is negative
X is the downrange horizontal distance
Vi is initial velocity
θ is the launch angle which is usually + but may be 0 or even negative.
all values are in MKS units and g= (-9.81 m/s/s)
in this problem
∆Y= ?
θ=59.3 deg
Vi=78.6 m/s
X = 26.0m
∆Y = 26.0*tan59.3 + ( ((-9.81)*26.0^2) / 2(78.6*cos59.3)^2 )
∆Y = 43.79 + (-6632) / 3221 = 41.7m
so
41.7 - 12.8 = 28.9m clearance