A projectile is shot at a velocity of # 8 m/s# and an angle of #pi/6 #. What is the projectile's peak height?

1 Answer
Dwight
Dec 17, 2016

Equations of motion are used to find an answer of 0.816 m

Explanation:

Use the equation of motion that reads:

#v^2# = #u^2# - #2gy#

Where #v# is the final vertical velocity, #u# is the initial vertical velocity and #y# is the change in altitude (the vertical displacement) of the projectile.

Now, the launch angle is #pi#/6, so #u# = 8 sin #pi/6# = 8 (0.5) = 4.0 #m/s#

"Reaches its greatest height" means #v# = 0, so putting it together, we get

0 = #4.0^2# - 2 (9.8)#y#

#y# = #16/ 19.6# = 0.816 m

Related questions
See all questions in Projectile Motion
Impact of this question
1379 views around the world
You can reuse this answer
Creative Commons License