If a projectile is shot at an angle of $(pi)/8$ and at a velocity of $21 m/s$ , when will it reach its maximum height?

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Answer 1 · 2018-04-01T15:39:18

The time is $=0.82s$

Resolving in the vertical direction $uarr^+$

The initial velocity is $u_0=(21)sin(1/8pi)ms^-1$

At the greatest height, $v=0ms^-1$

The acceleration due to gravity is $a=-g=-9.8ms^-2$

The time to reach the greatest height is $=ts$

Applying the equation of motion

$v=u+at=u- g t$

The time is $t=(v-u)/(-g)=(0-(21)sin(1/8pi))/(-9.8)=0.82s$

If a projectile is shot at an angle of (pi)/8(pi)/8 and at a velocity of 21 m/s21 m/s, when will it reach its maximum height?