A long jumper leaves the ground at an angleof 21.8 degrees to the horizontal at a speed of 10.6 m/s. What maximum height does he reach?
The long jumper leaves the ground at an angle of 21.8 degrees to the horizontal at a speed of 10.6 m/s.
There is a vertical acceleration acting downwards due to the gravitational pull of the Earth equal to 9.8 m/s^2. There is no horizontal acceleration that affects the path of the long jumper.
As the long jumper moves higher, the vertical component of his speed decreases and at the highest point it is equal to 0.
The vertical component of velocity at which the long jumper jumps is equal to 10.6*sin 21.8. If the velocity becomes 0 in t seconds we have 0 = 10.6*sin 21.8 - 9.8*t
=> t = 10.6*sin 21.8/9.8
The vertical distance traveled in this duration is (10.6*sin 21.8)^2/9.8 = 0.413 m
The maximum height of the long jumper is 0.413 m.
The jumper's horizontal velocity is 10.6m/s COS (21.8) but you don't need this.
His vertical velocity is 10.6m/s SIN (21.8)
First find the time he travels upward. t = (final velocity up - initial velocity up) / acceleration due to gravity. Final velocity going up is zero because he stops his vertical rise for a moment before falling back down. His acceleration going up is - 9.8 m/s^2 because he is slowing down.
So the time = [0 - 10.6m/s SIN (21.8)]/ - 9.8 m/s^2
Now you have his vertical velocity, his acceleration and the time going up.
vertical distance = (initial vertical velocity)t + (acceleration/2)t^2
Plop in the values, remembering that the acceleration is negative, and you will have the maximum vertical height he reaches.
The problem uses two equations: a= (v2 - v1)/t and d = v1t + 1/2 at^2