physics
posted by niyah .
a long jump athlete take off at 25 degree with the horizontal and achieves a jumping distance of 9.12 m. Calculate initial take off speed.

distanchoriz=vh*timeinair
but time in air can be found by
hf=hi+vy*time4.9t^2
or 0=0+vy*t4.9t^2
0=t(vy4.9) or time in air=vy/4.9
but vy=vsin25 and vh=vcos25
time in air=V*sin25/4.9
9.12=Vcos25*Vsin25/4.94.9(V *sin 25/4.9)^2
and you solve with V with some algebra and trig. 
10.8