
A ball P is dropped vertically and another ball Q is thrown horizontally with the same velocities from the same height and at the same time. If air resistance is neglected, then
A. Ball P reaches the ground first
B. Ball Q reaches the ground first
C. Both reach the ground at the same time
D. The respective masses of the two balls will decide the time
Answer
163.2k+ views
Hint:When a ball is dropped vertically and another ball is thrown horizontally with the same height, same velocities and same time, then initial velocity for both cases will be zero. Also for each case acceleration due to gravity, $g$ acts towards the ground. So, we can find out the time of reaching the ground by using the third equation of motion.
Formula used:
From the third equation of motion, we can write the formula:
Displacement,$s=ut+\dfrac{1}{2}g{{t}^{2}}$
Here $u=$Initial velocity
$t$ and $g$ is the time and acceleration due to gravity.
Complete step by step solution:
As both the balls were initially in rest position. Hence the value of $u$ is zero.
Thus, by putting this value into the equation the third of motion we can get,
$s=0\times t+\dfrac{1}{2}g{{t}^{2}}=\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow {{t}^{2}}=\dfrac{2s}{g}$
Equation for time required by ball to reach the ground is given by,
$t=\sqrt{\dfrac{2s}{g}}$
Here the displacement of the ball is the height from which it is thrown. Hence, $s = h$
So, $t=\sqrt{\dfrac{2h}{g}}$
The above equation signifies that the time taken by ball to reach the ground depends only on the height h from the ball is thrown irrespective of the way it is thrown horizontally and vertically. $g$ is a constant. As the height from which the ball is thrown is the same so the two balls will reach the ground at the same time.
Therefore, the correct option is C.
Note: When a ball is thrown vertically its kinetic energy keeps on decreasing and at maximum height velocity is zero. But potential energy keeps on increasing and at maximum height only potential energy will exist. Also when the same ball is dropped vertically the potential energy keeps on decreasing and kinetic energy increases gradually.
Formula used:
From the third equation of motion, we can write the formula:
Displacement,$s=ut+\dfrac{1}{2}g{{t}^{2}}$
Here $u=$Initial velocity
$t$ and $g$ is the time and acceleration due to gravity.
Complete step by step solution:
As both the balls were initially in rest position. Hence the value of $u$ is zero.
Thus, by putting this value into the equation the third of motion we can get,
$s=0\times t+\dfrac{1}{2}g{{t}^{2}}=\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow {{t}^{2}}=\dfrac{2s}{g}$
Equation for time required by ball to reach the ground is given by,
$t=\sqrt{\dfrac{2s}{g}}$
Here the displacement of the ball is the height from which it is thrown. Hence, $s = h$
So, $t=\sqrt{\dfrac{2h}{g}}$
The above equation signifies that the time taken by ball to reach the ground depends only on the height h from the ball is thrown irrespective of the way it is thrown horizontally and vertically. $g$ is a constant. As the height from which the ball is thrown is the same so the two balls will reach the ground at the same time.
Therefore, the correct option is C.
Note: When a ball is thrown vertically its kinetic energy keeps on decreasing and at maximum height velocity is zero. But potential energy keeps on increasing and at maximum height only potential energy will exist. Also when the same ball is dropped vertically the potential energy keeps on decreasing and kinetic energy increases gradually.
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