A projectile is shot from the ground at a velocity of $2 \frac{m}{s}$ $2 m/s$ at an angle of $\frac{π}{12}$ $pi/12$ . How long will it take for the projectile to land?

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Answer 1 · 2017-12-27T07:58:39

The time is $= 0.106 s$ $=0.106s$

Resolving in the vertical direction ${↑ ⏐ ⏐}^{+}$ $uarr^+$

The initial velocity is $u_{0} = 2 sin (\frac{1}{12} π) m s^{- 1}$ $u_0=2sin(1/12pi)ms^-1$

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

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

Applying the equation of motion

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

The time is $t = \frac{v - u}{- g} = \frac{0 - 2 sin (\frac{1}{12} π)}{- 9.8} = 0.053 s$ $t=(v-u)/(-g)=(0-2sin(1/12pi))/(-9.8)=0.053s$

The time taken to land is twice the time to reach the greatest height

$T = 2 t = 2 \cdot 0.053 = 0.106 s$ $T=2t=2*0.053=0.106s$

A projectile is shot from the ground at a velocity of 2ms2 m/s at an angle of π12pi/12. How long will it take for the projectile to land?