Time Of Flight

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A projectile is thrown with velocity u at an angle θ with the horizontal. The time of flight is:

RankGuru · Accepted Answer

1. **Vertical component of velocity**: $u_y = u \sin 	heta$.
2. **Peak Condition**: At the top of the trajectory, vertical velocity $v_y = 0$.
3. **Time to peak ($t_p$)**: $v_y = u_y - gt \implies 0 = u \sin 	heta - g t_p \implies t_p = \frac{u \sin 	heta}{g}$.
4. **Total flight ($T$)**: Assuming level ground, time to rise equals time to fall.
   - $T = 2 t_p = \frac{2 u \sin 	heta}{g}$.
5. **Result**: The time taken for the projectile to return to the initial horizontal level is **$2u \sin 	heta / g$**.

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More time of flight Practice Questions

A projectile is fired with speed 25 m/s at 45°. Find its time of flight (g = 10 m/s²).

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