Question

a ball is thrown vertically down with 5m/s. with what velocity should another ball be thrown down after 2 seconds so that it cam hit 1sy ball in 2 seconds​

Answer 1

Answer:

40 m/s

Explanation:

Veloctiy of 1st ball (u1) = 5 m/s

time taken by 1st ball(t1) = 2+2 = 4sec

Acceleration due to gravity = 10 m/s²

Distance travelled by 1st ball = S1

From the equation of motion

$\red{\boxed{\green{S = ut+\dfrac{1}{2}a{t}^{2}}}}$

$S_{1} = (5)(4)+\dfrac{1}{2}(10){(4)}^{2}$

$S</strong>_{1}<strong> = </strong>2<strong>0</strong><strong>+</strong>5<strong>(</strong><strong>1</strong><strong>6</strong><strong>)</strong><strong>$

$S_{1} = 20+80$

$S_{1} = 100m$

Now

time taken by second ball = 2 sec

Acceleration due to gravity = 10 m/s²

Distance travelled by second ball = S2

Acc. to the question, S1 = S2

velocity of second ball{u2} = ??

$S_{</strong><strong>2</strong><strong>} = </strong><strong>u</strong><strong>_</strong><strong>{</strong><strong>2</strong><strong>}</strong><strong>(</strong><strong>2</strong><strong>)+\dfrac{1}{2}(10){(</strong><strong>2</strong><strong>)}^{2}$

$S_{1} = 2u_{2}+5(4)$

$100 = 2u_{2}+20$

$100= 2(u_{2}+10)$

$u_{2}+10 = \dfrac{100}{2}$

$u_{2}+10 = 50$

$u_{2} = 50-10$

$u_{2} = 50-10$

$\pink{\boxed{\blue{u_{2} = 40 m/s}}}$