First slide

Vertical projection from ground

Question

From the top of a tower a body A is thrown up vertically with velocity u and another body B is thrown vertically down with the same velocity u. If $v_{A}$ and $v_{B}$ are their velocities when they reach the ground and $t_{A}$ and $t_{B}$ are their times of flight, then

Easy

A

$v_{A}$ = $v_{B}$ and $t_{A}$ = $t_{B}$
B

$v_{A}$ > $v_{B}$ and $t_{A}$ > $t_{B}$
C

$v_{A}$ = $v_{B}$ and $t_{A}$ > $t_{B}$
D

$v_{A}$ < $v_{B}$ and $t_{A}$ < $t_{B}$

Solution

Taking all upward physical quantities negative and downward physical quantities positive, for stone A, ${(v_{A})}^{2}$ = ${(- u)}^{2}$ + 2gh and for stone B, ${(v_{B})}^{2}$ = $u^{2}$ + 2gh

$v_{A}$ = $v_{B}$ stone A goes up, then comes down and then strikes the earth where as after projection stone B goes down and directly strikes the earth therefore $t_{A}$ > $t_{B}$

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