If a projectile is shot at an angle of $\frac{π}{2}$ $pi/2$ and at a velocity of $17 \frac{m}{s}$ $17 m/s$ , when will it reach its maximum height??

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Answer 1 · 2015-12-27T05:42:13

After $1.73 s$ $1.73"s"$

The launch angle = $\frac{π}{2}$ $pi/2$ radians which is $90^{\circ}$ $90^@$ which means a vertical launch.

$v = u + a t$ $v=u+at$

This becomes:

$0 = u - g t$ $0=u-"g"t$

$:.0=17-9.8t$

$:.t=17/9.8=1.73"s"$

If a projectile is shot at an angle of pi/2π2pi/2 and at a velocity of 17 m/s17ms17 m/s, when will it reach its maximum height??