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.

Answer 1

distanchoriz=vh*timeinair
but time in air can be found by
hf=hi+vy*time-4.9t^2
or 0=0+vy*t-4.9t^2
0=t(vy-4.9) or time in air=vy/4.9

but vy=vsin25 and vh=vcos25

time in air=V*sin25/4.9
9.12=Vcos25*Vsin25/4.9-4.9(V *sin 25/4.9)^2

and you solve with V with some algebra and trig.

Answer 2

10.8