JEE Main 2022 · 24 Jun · Shift 1 · Physics Q57

← JEE Main 2022 · 24 Jun · Shift 1

PhysicsInteger+4 / -1Question 57 of 90

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ______ s.

Your answer

Answer and worked solution

Official answer: 3

Equations (downward positive, height $h$ , speed $u$ ):
$h=-6u+\frac12g(36)$ and $h=1.5u+\frac12g(2.25)$ .

Eliminate $u$ . Multiply the second by 4 and add: $5h=18g+4.5g=22.5g$ , so $h=4.5g=\frac12g(9)$ .

Third ball. $h=\frac12gt_3^2$ , so $t_3=\sqrt{t_1t_2}=\sqrt{6\times1.5}=3$ s.

Answer: 3

Topics:Motion in a straight line

← Previous question Next question →

JEE Main 2022 · 24 Jun · Shift 1, Physics question 57