An Olympic long jumper is capable of jumping 8.3 m. Assume that his horizontal speed is 9.5 m/s as he leaves the ground. He was in the air for .87 seconds. How high does he go assuming that he lands upright, the same way he was when he left the ground?
8.3=viSinTheta*t you know vi, you know t, solve for theta.
you know it took .87/2 seconds to fall from max height.
h=1/2 g t^2 find h
To determine the maximum height reached by the Olympic long jumper, we can use the equations of motion. Since the long jumper is assumed to land upright, we can assume that he takes off and lands at the same height. Therefore, the maximum height reached can be found using the following equation:
h = (Vf^2 - Vi^2) / (2 * g)
where:
h = maximum height
Vf = final vertical velocity
Vi = initial vertical velocity
g = acceleration due to gravity
First, let's find the initial vertical velocity. We know that the jumper's horizontal speed is 9.5 m/s, and the time in the air is 0.87 seconds. Since there are no horizontal forces acting on the jumper during the flight, the horizontal speed remains constant.
Therefore, the horizontal displacement is given by:
d = Vx * t
where:
d = horizontal displacement
Vx = horizontal speed
t = time in the air
Substituting the given values:
d = 9.5 m/s * 0.87 s
d ≈ 8.265 m
Next, we can calculate the initial vertical velocity (Vi) using the horizontal displacement (d), which is equal to the horizontal speed (Vx) multiplied by the time (t) in the air.
Vi = d / t
Vi = 8.265 m / 0.87 s
Vi ≈ 9.50 m/s
Now, we can calculate the maximum height (h) reached by the long jumper.
h = (Vf^2 - Vi^2) / (2 * g)
Since the jumper lands upright, the final vertical velocity (Vf) is 0 m/s, as he will momentarily come to rest at the highest point.
Plugging in the values:
h = (0^2 - 9.50^2) / (2 * 9.8 m/s^2)
h = (-90.25) / 19.6
h ≈ -4.61 m
The negative sign indicates that the maximum height is below the take-off point. Therefore, the long jumper does not reach any significant height and lands slightly lower than his initial take-off height.