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.
3 answers
Your school SUBJECT is probably physics.
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
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
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?
1 answer