If a projectile is shot at an angle of (3pi)/8 and at a velocity of 7 m/s, when will it reach its maximum height?

1 Answer
Mar 13, 2018

The projectile will reach the maximum height at time t=(7/10).cos(pi/8) seconds after being thrown.

Explanation:

The initial velocity of the projectile is inclined at an angle theta with the horizontal.
Therefore the vertical motion which can be treated independently from the horizontal motion...will reach a point where its final velocity is zero.

and that instant is the position of maximum height.

The initial velocity component in the vertical direction is u * cos (pi/8).

So, we have

v = u * cos(pi/8) - g * t

where v is the final velocity, which is zero.

Therefore, the projectile will reach the maximum height at a

t = {u * cos( pi/8) } /g

Substituting the values of u and g, we have

t = (7/10.) cos (pi/8)