Horizontal problem:
U = 55 cos 10 = 54.16 m/s
It goes 80 meters horizontal at that speed
so
t = 80/54.16 = 1.48 seconds
Are you sure that is all you want to know?
I would ask you how high h was.
A projectile is fired from a height of h meters at a speed of 55m/s and at an angle of 10 degrees above the horizontal. The projectile hits a target at a horizontal distance of 80m from where it was fired and at a vertical height of 1.2m.
Find the time for which the projectile is in the air before it hits the target?
3 answers
Just in case that should come up
Vertical problem:
final height = h + Vi t - 4.9 t^2
1.2 = h + Vi t - 4.9 t^2
we know Vi and t from the horizontal problem.
Vertical problem:
final height = h + Vi t - 4.9 t^2
1.2 = h + Vi t - 4.9 t^2
we know Vi and t from the horizontal problem.
Oh, I never did Vi in fact
Vi = 55 sin 10 = 9.55 m/s
Vi = 55 sin 10 = 9.55 m/s
2 answers