A test rocket is launched by accelerating it along a 200.0 m incline at 1.86 m/s

Question

A test rocket is launched by accelerating it along a 200.0 m incline at 1.86 m/s^2 starting from rest at point A. The incline rises at 35.0 degrees 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.

Explanation / Answer

net acceleration (up along) = a =1.76 v^2 (launch) = 0 + 2 a s = 2 *1.76*200 = 704 v (launch) = 26.53 m/s height of launch = h(launch) = 200 * sin 35 = 114.72 m ---------------------------------- angle of lanuch = 35 deg position and velocity components after time (t) y - h = v sin 35 * t - 0.5 gt^2 ------ (1) x = v cos 35 * t ---- (2) Vx = dx/dt = v cos 35 ------(3) Vy = dy/dt = v sin 35 - gt ------ (4) --------------------------------------… at max height > Vy=0 = v sin 35 - gt t = v sin 35/g >> put in (1) h(max) = y = h + v sin 35 *[v sin 35/g] - 0.5g[v sin 35/g]^2 h(max) = h + v^2 sin^2 (35)/2g h(max) = 114.72 + [704*0.33/2*9.8] = 126.57 meter --------------------------------------… Range for time of flight (T) >>> putting y=0 >> or rocket hits the ground (1) >> 0 - h = v sin 35 * T - 0.5 g T^2 gT^2 - 2v sin 35 * T - 2h = 0 ------ (5) 9.8 T^2 - (30.43) T - 229.44 =0 solving >. T = 6.634 s (leaving -ve time) R = v cos 35 * T R = 26.53 [cos 35] *6.634 R = 144.17 meter ============ Range from point A (rest) = p + R where p = 200 cos 35 = 163.83 m R(t) = 308 meter

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