If a projectile is shot at an angle of $\frac{π}{8}$ $(pi)/8$ and at a velocity of $32 \frac{m}{s}$ $32 m/s$ , when will it reach its maximum height?

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If a projectile is shot at an angle of $\frac{π}{8}$ $(pi)/8$ and at a velocity of $32 \frac{m}{s}$ $32 m/s$ , when will it reach its maximum height?

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Answer 1 · 2016-05-10T09:30:12

$t = 1.24 sec$ $t=1.24 "sec"$

Explanation:

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$the vertical component of velocity is zero,if Object reaches its maximum height$ $"the vertical component of velocity is zero,if Object reaches its maximum height"$

$v_{y} = v_{i} \cdot sin α - g \cdot t$ $v_y=v_i*sin alpha-g*t$

$0 = 32 \cdot sin (\frac{π}{8}) - 9.81 \cdot t$ $0=32*sin(pi/8)-9.81*t$

$0 = 32 \cdot 0.38 - 9.81 \cdot t$ $0=32*0.38-9.81*t$

$0 = 12.16 - 9.81 \cdot t$ $0=12.16-9.81*t$

$9.81 \cdot t = 12.16$ $9.81*t=12.16$

$t = \frac{12.16}{9.81}$ $t=(12.16)/(9.81)$

$t = 1.24 sec$ $t=1.24 "sec"$

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If a projectile is shot at an angle of $\frac{π}{8}$ $(pi)/8$ and at a velocity of $32 \frac{m}{s}$ $32 m/s$ , when will it reach its maximum height?

1 Answer

Explanation:

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If a projectile is shot at an angle of (pi)/8π8(pi)/8 and at a velocity of 32 m/s32ms32 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 $\frac{π}{8}$ $(pi)/8$ and at a velocity of $32 \frac{m}{s}$ $32 m/s$ , when will it reach its maximum height?