Question

Two stones A and B are thrown with velocity 10 m/s and 20 m/s from the top of a tower horizontally if they reach the ground after times t1 and t2 respectively, then
1) t1 =2t2
2)t2 =2t1
3)t1=t2
4) (t1)^2 =(t2)^2
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Answer 1

Answer:

t₁ = t₂

Explanation:

Let the horizontal velocity  of the stones be $u_{1x}$ and $u_{2x}$. The vertical velocity of the stones will be same as equal to zero because stones are thrown horizontally.

Let the vertical initial velocity of both stones be $u_y=0$

Using the first equation of motion for first stone,

$h=u_yt+\frac{1}{2}g{t_1}^2\\\;\\h=0+\frac{1}{2}g{t_1}^2\\\;\\t_1^2=\frac{2h}{g}$

Using the second equation of motion for second stone

$h=u_yt+\frac{1}{2}g{t_2}^2\\\;\\h=0+\frac{1}{2}g{t_2}^2\\\;\\{t_2}^2=\frac{2h}{g}$

Thus

t₁² = t₂²

t₁ = t₂

Hence option 3) is correct

Answer 2

Answer:

Solution:

time of flight for a projectile is given by :- t

$= \sqrt{ \frac{2h}{g} }$

cuarly,t does not depend on initial horizontal velocity

as h=same for A & B (height of tower)

so,we get

$t₁= t₂$