please help... A bullet is fired from the top of a building of height 120m with a sped of 170m/s at an angle?

Question

of 37 degrees with the horizontal. The question is/are find the time it takes the bullet to reach the maximum height and the maximum height of the trajectory of the bullet and the speed of the bullet right befoe it hits the ground. PLease help im having a quiz tomorow and i need to know how to do these types of poblems

is the first one 11.45 seconds? if so u dont need to show the work

i ment 10.45seconds

I got 177m/s and -39.4 degrees for the next two if someone could confirm this that would be nice

Answer 1

find the component of the velocity

since it ask the time to reach the maximun height, we use Vy
Vy = sin(37)170 = 102.3m/s

Vf = at + Vi
when it reaches the maximun height, its final velocity is 0m/s
0m/s = (-9.8m/s^2)t + 102.3m/s
-102.3m/s = (-9.8m/s^2)t
t=10.44s

When the projectile return to the level of its initial position, (120m) its vertical velocity is -102.3m/s, the same vertical speed as it leaves.

Vf^2 = 2ad + Vi^2
Vf^2 = 2(-9.8m/s^2)(-120m) + (-102.3m/s)^2
Vf^2 = 2352 + 10465.29
Vf^2 = 12817.29
Vf = -113.21m/s^2 (This is vertical speed)

since you ask the actual speed, use trig
V = 113.21 / sin(37) = 188.11m/s