Asked by Anonymous

A 5.00 kg stone is thrown upward at 7.50 m/s at an angle of 51.0 degrees above the horizontal from the upper edge of a cliff, and it hits the ground 1.50 s later with no air drag. Find the magnitude of its velocity vector just as it reaches the ground.
Answered by Ms. Sue
Your school SUBJECT is probably physics.
Answered by Steve
The horizontal speed does not change. It remains at

7.5 cos51° = 4.72 m/s

The vertical speed is given by v = Vo -9.8t, so it ends up at

7.5 sin51° - 9.8*1.5 = -8.87 m/s

So, the final speed is √(4.72^2 + 8.87^2) = 10.05 m/s
Answered by Anonymous
A car traveling 88 km/h is 110 m behind a truck traveling 75 km/h. how long will it take the car to reach the truck?
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