Physics!!!!!!?

Question

A projectile is fired at 45.0° above the horizontal. Its initial speed is equal to 127.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile?


At what time after being fired does the projectile reach this maximum height?

Answer 1

A kinematics problem in two dimensions, except you aren't asked anything about the horizontal dimension. So start the problem when the projectile is fired and end when it reaches the top of the arc:

vy = voy - gt
h = voy t - 1/2 g t²

where

vy = 0
voy = ( 127.5 m/s ) ( sin 45 degrees )
g = 9.8 m/s²

Solve the first equation for the elapsed time t, then use that time in the second equation to solve for height h.

Answer 2

Initial speed of the projectile = 127.5 m/s
The inclination = 45°
Its vertical component = 127.5 x Sin 45°
Its horizontal component = 127.5 x Cos 45°
Acceleration due to gravity = 9.8 m/s²
v² - u² = 2 x g x h
0² - (127.5/√2)² = 2 x (-9.8) x h
- 8128.125 = - 19.6 x h
h = - 8128.125 / -19.6 = 414.7 m(max. ht reached)
....... ......... ........ =====================
v = u + g x t
0 = 127.5/√2 + (-9.8) x t
t x 9.8 = 127.5/√2 = 90.169
t = 90.169 / 9.8 = 9.2 sec after being fired the projectile
.... .......... =======
reaches the max. height.

Answer 3

Considering that this question is straight from a first year Physics class, you should probably use your kinematic equations your teacher gave you in class, you know, learn something, not ask on yahoo.