the vertical speed v = 20/√2 - 9.8t
so, v=0 at t=1.44
That's when it has stopped rising snd started to fall.
The flight time is just 2t
the horizontal speed is a constant 20/√2, so the range is just 20/√2 * 2t
going up, the velocity is (20/√2, 20/√2)
when it hits, it's the same speed, just going down, right?
If the soccer player kicks the ball with an initial velocity of 20 m/s at an angle of 45 degrees relative to the ground then what is the time it takes for the ball to reach the maximum height, what is the time of flight, what is the range, and what are the components of the velocity vector just before the ball hits the ground?
1 answer
1 answer