True: the time of flight of the two balls is the same
Explanation:
The ball that is dropped is in free fall, so its time of flight can be determined by using the suvat equation
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
[tex]s=h[/tex] is the vertical displacement, which is the height from which it was dropped
u = 0 is the initial vertical velocity
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
t is the time of flight
Solving for t,
[tex]t=\sqrt{\frac{2h}{g}}[/tex]
The ball thrown horizontally follows a projectile motion, which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
However, the time of flight is completely determined by the vertical motion only, therefore using the equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where u = 0 is the initial vertical velocity of the ball (which is zero, since it is thrown horizontally). Since the ball is thrown from the same height, its time of flight is still
[tex]t=\sqrt{\frac{2h}{g}}[/tex]
So, the two balls reach the grund at the same time.
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