Do it as a simple ballistic problem. From 0 to 140 seconds, the velocity at 140

Question

Do it as a simple ballistic problem. From 0 to 140 seconds, the velocity at 140 s is 2.3 km/s at 45 degree.( assume 45 degree from launch) Calculate the magnitude of acceleration in the vertical and horizontal. Calculate how far it travels horizontally and vertically during this time. 'Show calculations Assume the acceleration from 140 to 300 seconds is only do to gravity and g = 9.80 m/s2 starting from a velocity of 2.3 km/s at 45 degree. (assume 45 degree at 300 seconds) Calculate the magnitude of acceleration in the vertical and horizontal. Calculate how far it travels horizontally and vertically during this time.

Explanation / Answer

(a) Final horizontal velocity = Vx = Vsin(45) = 1.63 km/s = Vcos(45) = Vy = Final vertical velocity Time 140 sec Vertical acceleration, Ay = Vy/t = 1.63/140 = 11.6 m/s^2 = 1.63/140 = Vx/t = Ax = horizontal Acceleration Distance travelled horizontally = distance travelled vertically = .5*Ax*t^2 = 113.86 km (b) Magnitude of acceleration in horizontal = 0 , Magnitude of acceleration in vertical = g = 9.8 m/s^2 Horizontal distance travelled = Vx*160 = 260.8 km Vertical distance travelled = Vy*160 - .5*g*(160)^2 = 135.36 km

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