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
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
3 answers
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
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.
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.