A projectile is shot from the ground at a velocity of $2 m/s$ and at an angle of $(pi)/2$ . How long will it take for the projectile to land?

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Answer 1 · 2018-04-11T12:23:52

The time is $=0.41s$

Resolving in the vertical direction $uarr^+$

The initial velocity is $u=(2)sin(pi/2)=2ms^-1$

The acceleration due to gravity is $a=-g=-9.8ms^-2$

Apply the equation of motion

$s=ut+1/2at^2$

$s=0m$

Therefore,

$0=2t-1/2*9.8*t^2$

$4.9t^2-2t=0$

$t(4.9t-2)=0$

$t=0$ which is the initial conditions

$t=(2)/4.9=0.41s$

A projectile is shot from the ground at a velocity of 2 m/s2 m/s and at an angle of (pi)/2(pi)/2. How long will it take for the projectile to land?