Question

a rocket is launched vertically upwards from a cliff at a speed of 39.2ms-1. it hits the ground at the foot of the cliff 10 seconds later. a find the height of the cliff. b find the maximum height reached by the rocket. c find the time taken to reach the maximum height. d find the speed at which the rocket hits the ground. e find the time taken for the rocket to return to the same height as it was launched.

Answer 1

The height of the cliff is 1910 m. The maximum height reached by the rocket is 76.8 m and it takes 4 seconds to reach that height. The speed at which the rocket hits the ground is -141.2 m/s.

Step-by-step explanation:

To solve this problem, we will use the equations of motion for an object in free fall. We will assume that the acceleration due to gravity is -9.8 m/s2 (taking downward as negative).

a) To find the height of the cliff, we can use the equation: h = ut + (1/2)gt2. Plugging in the given values, we have h = (39.2 m/s)(10 s) + (1/2)(-9.8 m/s2)(10 s)2. Solving for h, we find that the height of the cliff is 1910 m.

b) To find the maximum height reached by the rocket, we can use the equation: v = u + gt, where v is the final velocity at the maximum height. At the maximum height, the velocity is 0 m/s, so we can solve for t. Plugging in the values, we have 0 = 39.2 m/s + (-9.8 m/s2)t. Solving for t, we find that the time taken to reach the maximum height is 4 seconds. To find the maximum height, we can use the equation: h = ut + (1/2)gt2. Plugging in the values, we have h = (39.2 m/s)(4 s) + (1/2)(-9.8 m/s2)(4 s)2. Solving for h, we find that the maximum height reached by the rocket is 76.8 m.

c) The time taken to reach the maximum height is 4 seconds, as found in part b.

d) To find the speed at which the rocket hits the ground, we can use the equation: v = u + gt, where u is the initial velocity and t is the time taken to hit the ground. Plugging in the values, we have v = 39.2 m/s + (-9.8 m/s2)(10 s). Solving for v, we find that the speed at which the rocket hits the ground is -141.2 m/s (taking downward as negative).

e) To find the time taken for the rocket to return to the same height as it was launched, we can use the equation: h = ut + (1/2)gt2. Plugging in the values for the initial height and initial velocity, we have 1910 m = (39.2 m/s)t + (1/2)(-9.8 m/s2)t2. Solving for t using the quadratic formula, we find that the time taken for the rocket to return to the same height is approximately 19.5 seconds.