A ball is projected up at an angle θ with horizontal from the top of a tower with speed v. It hits the ground at point A after time tA with speed vA . Now, this ball is projected at same angle and speed from the base of the tower (located at point P) and it hits ground at point B after time tB with speed vB. Then

Question

A ball is projected up at an angle θ with horizontal from the top of a tower with speed v. It hits the ground at point A after time tA with speed vA . Now, this ball is projected at same angle and speed from the base of the tower (located at point P) and it hits ground at point B after time tB with speed vB. Then (A)  PA=PB  (B) tA < tB (C) vA < vB (D) Ball A hits the ground at an angle (-θ) with horizontal

Tardigrade Admin · Accepted Answer

Consider, the two projectiles projected from

O

and

P

as shown in the figure.
Range of both the particles

R = \frac{v ^{2} s i n 2 θ}{g}

\Rightarrow O A^{'} = PB = R

Clearly, velocity at the points

A^{'}

and

B

will be same. Velocity of the ball following the trajectory OC'A'. Will increase after A' due to accelerated motion under gravity. Therefore, we can say that

v_{A} > v_{B}

$P A = PB$

$t_{A} < t_{B}$

$v_{A} < v_{B}$

Ball A hits the ground at an angle $(- θ)$ with horizontal

PA=PB

tA​<tB​

vA​<vB​

Ball A hits the ground at an angle (−θ) with horizontal

$P A = PB$

$t_{A} < t_{B}$

$v_{A} < v_{B}$

Ball A hits the ground at an angle $(- θ)$ with horizontal