A test rocket is launched by accelerating it along a 200.0-m incline at 1.74 m/s

Question

A test rocket is launched by accelerating it along a 200.0-m incline at 1.74 m/s^2 starting from rest at point A (the figure.) The incline rises at 35.0 degree above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find the maximum height above the ground that the rocket reaches. Express your answer in meters to three significant figures. Find the greatest horizontal range of the rocket beyond point A. Express your answer in meters to three significant figures.

Explanation / Answer

a) Height at end of ramp = h = dsin = 200m * sin35º = 114.7 m
and velocity at end of ramp = v = (2ad) = (2 * 1.74m/s² * 200m) = 26.4 m/s

Now the max height equation:
max. height H = h + (V·sin)² / (2g) = 114.7m + (26.4m/s*sin35º)² / 19.6m/s²
H = 126.4 m

b) I like the trajectory equation for this:
y = h + x·tan - g·x² / (2v²·cos²)
where y = height at x-value of interest = 0 m
and h = initial height = 114.7 m
and x = range of interest = ???
and = launch angle = 35º
and v = launch velocity = 26.4 m/s

Dropping units for ease,
0 = 114.7 + x*tan35 - 9.8x²/(2*26.4²*cos²35) = 114.7 + 0.7002*x - 0.01048x²

This quadratic has roots at x = -76.4 m ignore
and x = 143.2 m

Hope this helps!

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