A projectile is shot from the ground at a velocity of $6 m/s$ and at an angle of $(3pi)/4$ . How long will it take for the projectile to land?

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Answer 1 · 2015-12-26T09:18:38

$0.87"s"$

Use the equation of motion:

$v=u+at$

Considering the vertical component of velocity:

This becomes:

$0=vsintheta-"g"t$

$:.0=vsintheta-"g"t$

$:.t=(vsintheta)/g$

Converting radians to degrees:

$2pi=360^@$

$:.pi=360/2=180^@$

$:.(3pi)/4=(3xx180)/4=135^@=theta$

$t=(6sin((3pi)/(4)))/(9.8)$

$t=(6xx0.707)/9.8=0.433"s"$

This is the time to reach the zenith so the total time of flight will be $2xx0.433=0.866"s"$

A projectile is shot from the ground at a velocity of 6 m/s6 m/s and at an angle of (3pi)/4(3pi)/4. How long will it take for the projectile to land?