If a projectile is shot at an angle of $\frac{π}{2}$ $pi/2$ and at a velocity of $64 \frac{m}{s}$ $64 m/s$ , 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 $64 \frac{m}{s}$ $64 m/s$ , when will it reach its maximum height??

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Answer 1 · 2018-03-10T23:39:29

It reaches its highest point $6.5 s$ $6.5 s$ after launch.

Explanation:

$\frac{π}{2}$ $pi/2$ is straight up so we do not need to calculate a vertical component of the initial velocity. We can use the kinematic formula

$v = u + a \cdot t$ $v = u + a*t$

where $v = 0$ $v = 0$ because it stops at maximum height,
$u = 64 \frac{m}{s}$ $u = 64 m/s$ ,
and $a = - 9.8 \frac{m}{s^{2}}$ $a = -9.8 m/s^2$ which is the acceleration due to gravity, g. The value of a gets a minus sign because it is in the opposite direction of the initial velocity.

$v = u + a \cdot t$ $v = u + a*t$

$0 = 64 \frac{m}{s} - 9.8 \frac{m}{s^{2}} \cdot t$ $0 = 64 m/s - 9.8 m/s^2*t$

Solving for t,

$t = (64 cancel(m)/s)/(9.8 cancel(m)/s^2)$
(hmmm - I struggled with the formatter here. I wanted to strike out the power of 2 on the s, leaving just s. Would not work for me.)

$t = 64 /9.8 s = 6.53 s ~= 6.5 s$

I hope this helps,
Steve

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If a projectile is shot at an angle of $\frac{π}{2}$ $pi/2$ and at a velocity of $64 \frac{m}{s}$ $64 m/s$ , when will it reach its maximum height??

1 Answer

Explanation:

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If a projectile is shot at an angle of $\frac{π}{2}$ $pi/2$ and at a velocity of $64 \frac{m}{s}$ $64 m/s$ , when will it reach its maximum height??