A jumper in the long-jump goes into the jump with a speed of 10m/s at an angle of 30 degrees above the horizontal. Use g=10m/s.
a) how long in the air is the jumper before returning to the Earth?
b) How far does the jumper jump?
for a) I got .001s as the answer which logically seems wrong. I fond the vertical and horizontal component of velocity with:
v cos theda and v sin theda
(10)cos30 =8.66 and 10sin30 =5
then i found the v=sqrt(8.66^2 +5^2) and got v=9.99m/s
to find the time i used v=v_0+at
9.99=10+10t and then i got t=.001s
For b) I used x=x_0+vt and got .00866m. But my answers don't seem to be right. What am i doing wrong?
To find the correct answers for part (a) and part (b), let's go through the calculations step by step.
For part (a), we need to find the time the jumper is in the air before returning to the Earth.
1. Break down the initial velocity into its horizontal and vertical components:
- Horizontal component: v_x = v * cos(θ) = 10 m/s * cos(30°) = 8.66 m/s
- Vertical component: v_y = v * sin(θ) = 10 m/s * sin(30°) = 5 m/s
2. Since the jumper is returning to the same level, we can focus on the vertical component to find the time. Use the equation:
y = y_0 + v_y₀t + (1/2)gt^2
Here, y = 0 (as the jumper returns to the Earth surface), y_0 = 0 (initial height), v_y₀ = 5 m/s, and g = 10 m/s².
Substitute the given values and rearrange the equation to solve for t:
0 = 0 + 5t - (1/2) * 10 * t^2
0 = t(5 - 5t)
Apply the zero-product property:
t = 0 (no physical significance in this case) or 5 - 5t = 0
5t = 5
t = 1 s
Therefore, the jumper is in the air for 1 second before returning to the Earth.
For part (b), we need to find the horizontal distance covered by the jumper.
1. Use the equation:
x = x₀ + v_x₀ * t
Here, x₀ = 0 (initial horizontal position), v_x₀ = 8.66 m/s (horizontal component of initial velocity), and t = 1 s (from part (a)).
Substitute the given values and calculate:
x = 0 + 8.66 * 1
x = 8.66 m
Therefore, the jumper covers a horizontal distance of 8.66 meters.