A test rocket is launched by accelerating it along a 200.0-{\ m m} incline at 1.

Question

A test rocket is launched by accelerating it along a 200.0-{ m m} incline at 1.16{ m m/s^2} starting from rest at point A (the figure (Figure 1) .) The incline rises at 35.0 ^circ

a) Find the maximum height above the ground that the rocket reaches.
b) Find the greatest horizontal range of the rocket beyond point A.

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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