A long jumper leaves the ground at an angle of 17 degrees to the horizontal and at a speed of 9.43 m/s. How far does he jump? What maximum height does he reach?Answer in units of m. The...
A long jumper leaves the ground at an angle of 17 degrees to the horizontal and at a speed of 9.43 m/s. How far does he jump? What maximum height does he reach?
Answer in units of m. The acceleration due to gravity is 9.8 m/s2.
To solve this, you need to know how much of his velocity is horizontal and how much is vertical.
Construct a right triangle, starting at the origin, with an angle of 17 degrees, and a hypotenuse of 9.43 m/s. Then use sine and cosine to find the x- and y-values for velocity.
Sin 17 = y/9.43 m/s and y = 2.76 m/s vertical
cos 17 = x/9.43 m/s and x = 9.02 m/s
In the vertical direction the jumper has an initial velocity of 2.76 m/s but is slowing to a stop at the highest point because of the downward acceleration of gravity.
Since Vf = Vi + gt, where Vf is the final velocity which is zero; Vi is the initial velocity which is 2.76 m/s and g is a -9.8 m/s/s if up is defined as the positive direction.
0 = 2.76 m/s + (-9.8 m/s/s) t
t = .282 seconds up + .282 seconds down for a total time in the air of 0.563 s,
The entire time the jumper is in the air he is moving horizontally at 9.02 m/s. Since d = vt, d = 9.02 * 0.563 = 5.08 m jump.
The maximum height in the y-direction is given by:
delta y = Vit + 1/2 gt^2
delta y = 2.76 m/s * 0.282 s + 1/2 (-9.8 m/s/s)(0.282^2)
delta y = 0.39 m high