let his initial velocity be V.
Then the initiualvertical velocity was V*sin31. Initial horizontal velocity= V*cos31
time in air:
t=8.30m/Vcos31
In the vertical:
hf=hi+vi't-4.9t^2
0=vSin31*8.30m/Vcos31-4.9(8.30m/Vcos31)^2
so, solve for V, a little albebra involved.
At a track and field meet, the best long jump is measured as 8.30 m. The jumper took off at an angle of 31 ° to the horizontal. What was the jumper’s initial speed?
2 answers
or, more directly, recalling that the range is
R = v^2/g sin2θ
just solve for v in
v^2/9.8 * sin62° = 8.3
R = v^2/g sin2θ
just solve for v in
v^2/9.8 * sin62° = 8.3
3 answers