Question
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?
Find the time for which the projectile is in the air before it hits the target?
Answers
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.
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.
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