If a projectile is shot at a velocity of $17 \frac{m}{s}$ $17 m/s$ and an angle of $\frac{π}{4}$ $pi/4$ , how far will the projectile travel before landing?

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If a projectile is shot at a velocity of $17 \frac{m}{s}$ $17 m/s$ and an angle of $\frac{π}{4}$ $pi/4$ , how far will the projectile travel before landing?

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Answer 1 · 2015-12-31T13:33:29

The projectile travels $29.45 m$ $29.45m$

Explanation:

Consider the vertical equation $s (y) = v t - \frac{1}{2} a t^{2}$ $s(y)=vt-1/2at^2$ and assume it equal to 0.
$s (y) = t (v - \frac{1}{2} a t) = 0$ $s(y)=t(v-1/2at)=0$ so $t = 2 \frac{v}{g} = 2 v \frac{sin 45}{g} = 2.45 s$ $t=2v/g=2vsin45/g=2.45s$
Use this value in the horizontal equation $s (x) = v t = v cos 45 \cdot t = v cos 45 \cdot 2.45 = 29.45 m$ $s(x)=vt=vcos45*t=vcos45*2.45=29.45m$

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If a projectile is shot at a velocity of $17 \frac{m}{s}$ $17 m/s$ and an angle of $\frac{π}{4}$ $pi/4$ , how far will the projectile travel before landing?

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If a projectile is shot at a velocity of 17ms17 m/s and an angle of π4pi/4, how far will the projectile travel before landing?

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If a projectile is shot at a velocity of $17 \frac{m}{s}$ $17 m/s$ and an angle of $\frac{π}{4}$ $pi/4$ , how far will the projectile travel before landing?