If a projectile is shot at an angle of pi/6 and at a velocity of 15 m/s, when will it reach its maximum height??

2 Answers
George C.
Jan 3, 2016

Assuming g = 9.8ms^(-2) it will reach its maximum height after 0.765 seconds.

Explanation:

The vertical component of the velocity is 15*sin(pi/6) = 7.5 ms^(-1)

The gravitational acceleration is 9.8ms^(-2).

So it will take 7.5/9.8 ~~ 0.765 seconds before the vertical component of the velocity is zero.

Max
Jan 4, 2016

0.76s

Explanation:

I always find a diagram helpful:
My own image
I'm going to make the assumption for this question that we can ignore air resistance.
I am also defining right and up as positive directions.
This allows us to use the constant acceleration equations.

It should also be noted that at the maximum height, vertical velocity is zero.

This equation seems ideal for this problem, as it contains all the variables that we have values for and the one we wish to solve.

v_y=u_y+a_yt

Let's see what values we have:

v_y=0
u_y=15sin(pi/6)=7.5
a_y=-g=-9.81
t=?

Plugging them in:

0=7.5-9.81t

t=7.5/9.81

t=0.76452...s

t=0.76s (2sf)

Related questions
See all questions in Falling Objects
Impact of this question
1909 views around the world
You can reuse this answer
Creative Commons License