If a projectile is shot at an angle of $\frac{π}{2}$ $pi/2$ and at a velocity of $37 \frac{m}{s}$ $37 m/s$ , when will it reach its maximum height??

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If a projectile is shot at an angle of $\frac{π}{2}$ $pi/2$ and at a velocity of $37 \frac{m}{s}$ $37 m/s$ , when will it reach its maximum height??

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Answer 1 · 2017-01-05T13:08:09

We need to find the instant at which the projectile's vertical component of velocity is zero.

Explanation:

The vertical component of the velocity at time $t$ $t$ is given by $v (t) = u - g t$ $v(t)=u-g t$ , where $u$ $u$ is the projectile's initial vertical component of velocity and $g$ $g$ is the gravitational acceleration constant.

So we are required to solve $0 = u - g t$ $0=u-g t$ , which gives $t = \frac{u}{g}$ $t=u/g$

Now, $u = 37 sin (\frac{π}{2})$ $u=37 sin (pi/2)$ m/s and we take $g = 10$ $g=10mbox{}$ m/s $^{2}$ $\mbox{}^2$ . So the final answer is $t = \frac{37}{10} = 3.7$ $t=37/10=3.7$ s.

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If a projectile is shot at an angle of $\frac{π}{2}$ $pi/2$ and at a velocity of $37 \frac{m}{s}$ $37 m/s$ , when will it reach its maximum height??

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Explanation:

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If a projectile is shot at an angle of $\frac{π}{2}$ $pi/2$ and at a velocity of $37 \frac{m}{s}$ $37 m/s$ , when will it reach its maximum height??