A projectile is shot at a velocity of 23 m/s and an angle of pi/4 . What is the projectile's peak height?

1 Answer
Apr 9, 2018

To answer this the formulas u_y = usin(theta), v_y = u_y + a_yt and y = u_yt + (a_yt^2)/2 can be used.

Explanation:

First we want vertical component of velocity
u_y = usin(theta)
u_y = 23sin(pi/4)
u_y = 23/sqrt(2)

Second we find the time taken to reach a vertical velocity of 0 (this is its peak height).
v_y = u_y + a_yt
0 = 23/sqrt(2) -9.8/t
t =9.8xxsqrt(2)/23
(Assuming the projectile is on Earth a_y = -9.8 )

Finally we can find the peak height
y = u_yt + (a_yt^2)/2
y = 23/sqrt(2) xx 9.8 xx sqrt(2)/23 - 9.8 xx (9.8 xx sqrt(2)/23)^2/2

y = 9.8 - 9.8^3/23^2
y = 8.02 m