A rocket, initially at rest, is fired horizontally with a horizontal acceleration of 12 m/s2. If the rocket is fired from a cliff 80 meters high, what is the distance in meters between the landing location and the bottom of the cliff? (Note: use g=10 m/s2).
Question
A rocket, initially at rest, is fired horizontally with a horizontal acceleration of 12 m/s2. If the rocket is fired from a cliff 80 meters high, what is the distance in meters between the landing location and the bottom of the cliff? (Note: use g=10 m/s2).
Solution
To solve this problem, we need to find two things: the time it takes for the rocket to hit the ground (vertical motion) and the horizontal distance it travels in that time (horizontal motion).
Step 1: Determine the time of flight The rocket is fired from a cliff, so it's essentially in free fall once it's launched. We can use the equation of motion to find the time it takes for the rocket to hit the ground:
h = 0.5 * g * t^2
Where: h = height from which the rocket is launched = 80 m g = acceleration due to gravity = 10 m/s^2 t = time of flight
Rearranging the equation to solve for t gives:
t = sqrt((2h)/g) = sqrt((280)/10) = sqrt(16) = 4 seconds
So, the rocket will be in the air for 4 seconds.
Step 2: Determine the horizontal distance The rocket is accelerating horizontally, so we can use the equation of motion:
d = v_it + 0.5a*t^2
Where: d = horizontal distance v_i = initial velocity = 0 (since the rocket is initially at rest) a = horizontal acceleration = 12 m/s^2 t = time of flight = 4 seconds
Substituting the known values gives:
d = 04 + 0.5124^2 = 0 + 0.512*16 = 96 meters
So, the distance between the landing location and the bottom of the cliff is 96 meters.
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