If a projectile is shot at an angle of $\frac{π}{4}$ $pi/4$ and at a velocity of $16 \frac{m}{s}$ $16 m/s$ , when will it reach its maximum height??

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Answer 1 · 2016-01-17T11:57:45

1.15s

A projectile reaches its maximum height when vertical velocity is zero.

Initial velocity, u = $16 cos (\frac{π}{4}) = 16 \cdot \frac{\sqrt{2}}{2} = 8 \sqrt{2}$ $16cos(pi/4)=16*sqrt2/2=8sqrt2$

Taking acceleration as $a = - 9.8 m s^{- 1}$ $a=-9.8ms^-1$ then:

$v = u + a t$ $v = u + at$
$0 = 8 \sqrt{2} - 9.8 t$ $0=8sqrt2-9.8t$
$t = \frac{8 \sqrt{2}}{9.8} = 1.15 s$ $t=(8sqrt2)/9.8=1.15s$ (3sf)

If a projectile is shot at an angle of π4pi/4 and at a velocity of 16ms16 m/s, when will it reach its maximum height??