Need help with short projectile motion question

a ball is thrown up off the roof of a school at an angle of 55 degrees from the horizontal and at a velocity of 25m/s.

how long before it hits the ground if it land 95meters from the base of the building?

at what horizontal and vertical velocity does it hit the ground

The horizontal velocity component remains at this value during flight:

Vx = 25 cos 55 = 14.34 m/s.
The vertical velocity component initially is
Vyo = 25 sin 55 = 20.48 m/s.
The time to hit the ground is
T = 95/Vx = 6.63 s

When it hits the ground, total kinetic energy increases so that
(Vy)^2 - (Vyo)^2 = 2 g H
where H is the height of the cliff

i thought T was the total time

dont you have to subtract it by the time of the angle to find the time of the fall

There was an omission in my previous answer. Since you were not given the height of the building, you need another method to compute the final vertical velocity component, Vy. Use this:

Vy = Vyo - g T = 20.48 - 65.04
= -44.46 m/s

I believe the method used to compute the time of flight, T, was correct.

The height of the building can be computed from the available information and
Vy,final^2 - Vyo^2 = 2 g H
= 1566 m^2/s^2
H = 79.8 m

That is a suspiciously large height for a school building.

To solve this projectile motion question, we'll need to break it down into two components: the horizontal motion and the vertical motion.

First, let's calculate the time it takes for the ball to hit the ground. Since the horizontal motion is unaffected by gravity, we can focus on the vertical motion. The initial vertical velocity (Vi) can be calculated using the launch angle (θ) and the initial velocity (V).

Vi = V * sin(θ)
Vi = 25 * sin(55°)
Vi ≈ 21.34 m/s

Using the equation of motion for free fall, we can find the time (t) it takes for the ball to hit the ground:

d = Vi * t + (1/2) * g * t^2
95 = 21.34 * t + (1/2) * 9.8 * t^2

Simplifying the equation gives us a quadratic equation:
4.9t^2 + 21.34t - 95 = 0

To solve the quadratic equation, you can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

Using the values from the equation, we have:
a = 4.9, b = 21.34, c = -95

Solving for t gives us:
t = (-21.34 ± √(21.34^2 - 4 * 4.9 * -95)) / (2 * 4.9)

Calculating the equation will give us two possible values for t. We'll choose the positive value since time cannot be negative:

t ≈ 3.84 seconds

So, it takes approximately 3.84 seconds for the ball to hit the ground.

To find the horizontal and vertical velocities at impact, we can use the following formulas:

Vfx = V * cos(θ)
Vfy = Vi + g * t

Using the given values and calculated time:
Vfx = 25 * cos(55°)
Vfx ≈ 14.32 m/s

Vfy = 21.34 + 9.8 * 3.84
Vfy ≈ -19.26 m/s

The negative sign indicates that the vertical velocity is directed downwards.

Therefore, the ball hits the ground with a horizontal velocity of approximately 14.32 m/s and a vertical velocity of approximately -19.26 m/s.