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?

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.

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.

Oh, I never did Vi in fact

Vi = 55 sin 10 = 9.55 m/s

To find the time for which the projectile is in the air before it hits the target, we can use the kinematic equations of motion in both the horizontal and vertical directions.

First, let's analyze the horizontal motion of the projectile. Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion. We can use the formula:

Horizontal distance = horizontal velocity × time

Given that the horizontal distance is 80m and the horizontal velocity is 55m/s, we can rearrange the formula to solve for time:

time = Horizontal distance / horizontal velocity

time = 80m / 55m/s

time ≈ 1.45 seconds

Therefore, the projectile is in the air for approximately 1.45 seconds before it hits the target.