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
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?
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