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?
time in air:
hfinal=hinitial +vyi*t -1/2 g t^2
hfinal, hinitial are zero, solve for t.
(vyi= 10sin30)
How far does he go: right equation
d= vix*t = 10cos20*time
In part a), you made a small calculation error when finding the initial vertical component of velocity. The correct calculation should be:
10sin30 = 5m/s (rounded to two decimal places)
Therefore, the correct calculation for finding the time in the air would be:
0 = 5 - 10t
t = 0.5s
So, the jumper is in the air for 0.5 seconds before returning to the Earth.
In part b), you made a mistake in the equation for finding the horizontal distance. The correct equation should be:
x = x_0 + vt
where x_0 is the initial horizontal position, which is typically assumed to be zero in this type of problem.
x = 0 + (10*cos30)*0.5
x = 8.66 * 0.5
x = 4.33m
So, the jumper jumps a horizontal distance of 4.33 meters.
To determine what may have gone wrong in your calculations for both parts (a) and (b), let's go through the steps again:
(a) Calculating the time of flight:
In your calculations, you correctly identified the vertical and horizontal components of the initial velocity: vy = 10 * sin(30) = 5 m/s and vx = 10 * cos(30) = 8.66 m/s. However, the equation you used to find the time is incorrect. Instead of v = v0 + at, you need to use the equation y = y0 + v0y*t + (1/2)*gt^2 to determine the time of flight. In this case, the final vertical displacement is zero since the jumper starts and ends at the same height. So, the equation becomes: 0 = 0 + (5)t + (1/2)(-10)t^2.
Solving this quadratic equation will give you two solutions: t = 0 and t = 1. The solution t = 0 represents the initial time when the jumper is about to take off. Therefore, the actual time of flight is t = 1 second.
(b) Calculating the horizontal distance:
To find the horizontal distance covered by the jumper, you correctly used the equation x = x0 + vt. However, you used the incorrect value of vx. The horizontal component of the velocity remains constant throughout the motion, so you should use its initial value, which is vx = 8.66 m/s.
Plug in the values in the equation: x = 0 + (8.66) * 1 = 8.66 m.
So, the jumper will be in the air for 1 second and will cover a horizontal distance of 8.66 meters.
By correcting the equation used for finding the time and using the correct initial horizontal velocity, you should get the correct answers.