A projectile is shot from the ground at a velocity of $24 m/s$ and at an angle of $(2pi)/3$ . How long will it take for the projectile to land?

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Answer 1 · 2016-01-21T08:24:55

$4.2416 s$

Data:-
Initial velocity $=v_0=24m/s$

Angle of throwing $=theta=(2pi)/3$

Time of flight $=T=??$

Acceleration due to gravity $=g=9.8m/s^2$

Sol:-

We know that:

$T=(2v_0sintheta)/g$

$implies T=(2*24*Sin((2pi)/3))/9.8=(48*0.866)/9.8=4.2416$

$implies T=4.2416s$

Hence the projectile will land at $4.2416 s$ approximately.

A projectile is shot from the ground at a velocity of 24 m/s24 m/s and at an angle of (2pi)/3(2pi)/3. How long will it take for the projectile to land?