A projectile is shot at a velocity of 4ms and an angle of π12. What is the projectile's maximum height?

1 Answer
Jan 14, 2016

h=(4×0.25882)22×9.81m
0.05463m

Explanation:

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Maximum height is reached because of sinθ component of the velocity.
Use the formula
v2u2=2gh
Given v=0,u=4.sin(π12); as θ=π12
Take g=9.81ms2
Since gravity is acting against the velocity so it is deceleration and sign is used in front of g
Substituting various values
h=(4.sin(π12))22×9.81m
Use tables to substitute the value of
(sin of given angle) = 0.25882