A projectile is shot at a velocity of 2 m/s and an angle of pi/6 . What is the projectile's peak height?

1 Answer
Mar 1, 2016

h_p~=0,051 m

Explanation:

"The projectile's peak height can be calculated using the formula:"
h_p=(v_i^2*sin^2 alpha)/(2*g)
v_i" :initial velocity of object"
alpha" :angle of projectile"
g" : acceleration of gravity"
alpha=pi/6*180/pi=30^o
sin 30=0,5
sin^2 30=0,25
h_p=(2^2*0,25)/(2*9,81)
h_p=(2*0,25)/(9,81)
h_p~=0,051 m