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

Answer:

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)#