Please show me the step by step solution to this problem A test rocket is launch

Question

Please show me the step by step solution to this problem

A test rocket is launched by accelerating it along a 200.0 m incline at 1.78 m/s2 starting from rest at point A. The incline rises at 35.0 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). a) Find the maximum height above the ground that the rocket reaches b) Find the greatest horizontal range of the rocket beyond point A (point A is located at the base of the incline)

Explanation / Answer

Given length of the inclined plane, is s = 200 m Angle of inclination , = 35^0 Acceleration of the rocket is ,a = 1.78 m/s^2 Time taken to accelerate along the incline is                            s = 1/2 at1^2                    200 m = 0.5 * 1.78 m/s^2 *t1^2                            t1 = 15 s The velocity of the rocket as it leaves the incline is                          v = at1 = 1.78 m/s^2 * 15s                           v = 26.7 m/s The horizontal component of velocity is , vxo = 26.7 m/s cos 35 =21.87 m/s The vertical component of velocity is , vyo = 26.7 m/s sin 35 =15.31 m/s The height of the incline is , h = 200 sin35 = 114.7 m The length of the base of the incline is , l = 200 cos35 = 163.83 m --------------------------------------------------------------------------------- Time taken to reach maximum height from the incline is t2 = vyo/g                                                       t2 = (15.31 m/s) /9.8 m/s^2 = 1.5 s The maximum height reached above the incline is H = 1/2 gt2^2                                                       H' = 0.5 *9.8 m/s^2 *(1.5 s)^2 = 11m Total time to reach the reach the launching point from the incline is t = 2 *1.5 s = 3.0s Horizontal distance travelled from the incline is ,x = 21.87 m/s * 3.0 s = 65.6 m ---------------------------------------------------------------------------------------- The maximum height above the ground H = h+H' = 114.7 m +11 m = 125.7 m The maximum horizontal range is R = x +l = 65.6 m +163.83 m = 229.43 m The vertical component of velocity is , vyo = 26.7 m/s sin 35 =15.31 m/s The height of the incline is , h = 200 sin35 = 114.7 m The length of the base of the incline is , l = 200 cos35 = 163.83 m --------------------------------------------------------------------------------- Time taken to reach maximum height from the incline is t2 = vyo/g                                                       t2 = (15.31 m/s) /9.8 m/s^2 = 1.5 s The maximum height reached above the incline is H = 1/2 gt2^2                                                       H' = 0.5 *9.8 m/s^2 *(1.5 s)^2 = 11m Total time to reach the reach the launching point from the incline is t = 2 *1.5 s = 3.0s Horizontal distance travelled from the incline is ,x = 21.87 m/s * 3.0 s = 65.6 m ---------------------------------------------------------------------------------------- The maximum height above the ground H = h+H' = 114.7 m +11 m = 125.7 m The maximum horizontal range is R = x +l = 65.6 m +163.83 m = 229.43 m

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