hf=hi+vi*t-4.9t^2 vi= 37.0*sin30.6 hf=8.9 hi=0 solve for time t.
YOu get two solutions, take the largest time, it is going down then.
Highest point? Vvertical=0
Vv=0=vi*t-4.9t^2 vi = 37*sin30.6
solve for time t.
then horizontal distce= 37cos30.6*time
A projectile is launched from ground level at 37.0 m/s at an angle of 30.6 ° above horizontal. Use the launch point as the origin of your coordinate system.
a)How much time elapses before the projectile is at a point 8.9 m above the ground and heading downwards toward the ground?
b)How far downrange (the horizontal distance from the origin) was the projectile when it reached the highest point in its flight?
1 answer
1 answer