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Zero Launch Angle - Car Drives Off a Cliff A car drives straight off the edge...

Zero Launch Angle - Car Drives Off a Cliff A car drives straight off the edge of a cliff that is h =40 m high. The police at the scene of the accident note that the point of impact is d =86 m from the base of the cliff. Here the initial speed v0 is unknown. Ignore air resistance. The magnitude of the gravitational acceleration is 9.8 m/s2. Choose the RIGHT as positive x-direction. Choose UPWARD as positive y-direction.

A. In this problem, even though the initial speed v0 is unknown, you can still find v0y, the y component of the initial velocity. What is v0y (in m/s) ?

B.How long (in seconds) is the car in the air before hitting the ground?

C. How fast (in m/s) was the car traveling when it leaves the cliff?

D. What is the vertical component of the velocity (in m/s) just before the car hits the ground? Pay attention to the direction (the sign).

E. What is the magnitude of the velocity (in m/s) (including both the horizontal and vertical components) of the car just before it hits the ground?

F. What is the direction of the velocity of the car just before it hits the ground?

Report the direction by an angle COUNTERCLOCKWISE from the +x axis.

G. Assume the car’s new speed leaving the cliff is 66.0 mph but the height of the cliff is unchanged, still 40 m.

H. What is the new speed in m/s? 1mile = 1609 m, 1 hour = 60 minutes, 1 minute = 60 seconds

m/s

I. what is the time interval (in seconds) from the car leaving the cliff to it hitting the ground?

J.What would be the horizontal distance d (in meters) from the base of the cliff to the impact point?

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