If a projectile is shot at an angle of $pi/4$ and at a velocity of $12 m/s$ , when will it reach its maximum height??

Question

1400 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-03T13:04:16

The time is $=0.87s$

Resolving in the vertical direction $uarr^+$

The initial velocity is $u_0=(12)sin(1/4pi)ms^-1$

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

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

At the maximum height $v=0ms^-1$

Applying the equation of motion

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

The time is $t=(v-u)/(-g)=(0-(12)sin(1/4pi))/(-9.8)=0.87s$

If a projectile is shot at an angle of pi/4pi/4 and at a velocity of 12 m/s12 m/s, when will it reach its maximum height??