Question

Balls 1 and 2 are each thrown horizontally from the same height above level ground, but ball 2 has a greater initial velocity after leaving the thrower’s hand. If air resistance is negligible, how do the accelerations of the balls and the times it takes them to hit the ground compare?

Answer 1

Answer:

Acceleration and time will be same for both balls.

Explanation:

Given that ball is thrown horizontally

Lets take both balls are at a height of h from ground level.

Lets take initial velocity of ball 1 = u

Lets take initial velocity of ball 2 = v

Given that v>u.

Initially the component of velocity in vertical direction is zero.

so

$h=\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h}{g}}$

So time to reach at ground will be same for both the balls.

We know that acceleration due to gravity is also same for both the balls.Acceleration for both ball will be g$m/s^2$.

Answer 2

The acceleration and time of motion for the two balls is the same since they have zero initial vertical velocity.

• Let the initial horizontal velocity of Ball 1 = v
• then, the initial horizontal velocity of Ball 2 = 2v
• the height traveled by the two balls, = h

The vertical height traveled by the balls is calculated as follows;

$h = v_y_0t + \frac{1}{2} gt^2$

For the first ball;

$h = 0 + \frac{1}{2} gt^2\\\\2h = gt^2\\\\g = \frac{2h}{t^2} \\\\t= \sqrt{\frac{2h}{g} }$

For the second ball;

$h = 0 + \frac{1}{2} gt^2\\\\2h = gt^2\\\\g = \frac{2h}{t^2} \\\\t= \sqrt{\frac{2h}{g} }$

Thus, we can conclude that the acceleration and time of motion for the two balls is the same since they have zero initial vertical velocity.

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