Vectors and Projectiles - Mission VP10 Detailed Help

A projectile is launched from the ground at an angle to the horizontal and subsequently hits the ground at the same initial height. The projectile requires 3.0 seconds (or 5.0 seconds) to reach the peak of its trajectory. Which of the following are true of the projectile?


 
This question pertains to an angle-launched projectile. Such a projectile rises vertically towards its peak, turns around, and falls vertically from its peak back to the original height. This rising and falling motion is accompanied by a constant speed motion in the horizontal direction. This horizontal motion is independent of the vertical motion and irrelevant to the question.
 
The time required for a projectile to rise to the peak is equal to the time for the projectile to fall from its peak. As such, if you wish to find the total time of flight, simply take the time to the peak and double it.
 
The time for the upward rise to the peak is dependent solely upon the original vertical velocity. The greater the original vertical velocity, the more time required to rise to the peak. The original vertical velocity value can be calculated using a kinematic equation (see Formula Frenzy section). The vertical velocity at the peak (vfy) is 0 m/s. The vertical acceleration (ay) is the acceleration of gravity (approximately -10 m/s/s). And the time (t) is the time required to travel from ground level to the peak position.


 
The final vertical velocity (vfy) of an object after a certain time (t) is related to the vertical acceleration (ay) and the original vertical velocity (voy) by the kinematic equation:

vfy= voy+ ay• t


 

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