Question

A ball is thrown up with a velocity 9.8 m/s. After 1 second another ball is thrown up with a
velocity of 19.6 m/s. What will be the distance between them 1 sec after the second ball is
1) 49m
2) 9.8 m
3) 14.7 m
4) 19.6m​

Answer 1

Answer:

Correct option

3) 14.7 m

Step by step explanations :

given that,

A ball is thrown up with a velocity

9.8 m/s.

here,

initial velocity of the ball = 9.8 m/s

also,

given that,

After 1 second another ball is thrown up with a velocity of 19.6 m/s.

here,

initial velocity of another ball = 19.6 m/s

given time interval = 1 s

now,

to find,

the distance between them 1 sec after the second ball is thrown,

here,

distance travelled by the 1st ball

here, time(t) = 2 seconds

initial velocity(u) = 9.8 m/s

gravitational acceleration(g) = -9.8 m/s²

so,

by the gravitational equation of motion,

h = ut + ½gt²

where,

h = distance travelled in given time interval,

putting the values,

h = 9.8(2) + ½(-9.8)(2)(2)

= 19.6 - 19.6

= 0 m

now,

distance travelled for the another ball,

initial velocity = 19.6 m/s

gravitational acceleration(g) = -9.8 m/s²

time taken(t) = 1 s

h = ut + ½gt²

putting the values,

h = 19.6(1) + ½(-9.8)(1)(1)

= 19.6 - 4.9

= 14.7 m

so,

distance between them,

= difference of distance travelled

= 14.7 - 0

= 14.7 m

so,

distance between them = 14.7 m

________________

Correct option

3) 14.7 m

Answer 2

Answer:-

H = 14.7m

Option → c

Given :-

A ball is thrown with velocity

9.8 m/s.

After 1 s another ball is thrown with a velocity of 19.6 m/s.

To find :-

The distance between them after 1 s.

Solution :-

In case 1:-

U = 9.8 m/s.

g = - 10m/s²

v = 0 m/s

Now, time is unknown so we use 3rd equation of motion.

★ $\boxed{\sf{h_1 = ut + \dfrac{1}{2}at^2}}$

Now put the given value,

→$h_1 = 9.8 \times 1 -\dfrac{1}{2} \times 9.8\times 1^2$

$h_1 = 9.8 - \dfrac{1}{2}\times 9.8$

→$h_1 =9.8 -4. 9$

→$h_1 = 4.9 m$

Now,

Case 2:-

u = 19.6 m/s

t = 1 s

g = -9.8 m/s²

Height cover by 2 nd ball is :-

→$h_2 = ut - \dfrac{1}{2}gt^2$

→$h_2 = 19.6 \times 2 \times 1 - \dfrac{1}{2}\times 9.8 \times (2) ^2$

→$h_2 = 39.2 - \dfrac{1}{2}\times 9.8 \times 4$

→$h_2 = 39.2 - \dfrac{1}{2}\times 9.8$

→$h_2 = 39.2 - 19.6m[\tex]</p><p></p><p>→[tex] h_2 = 19.6 m$

Now, the distance between them after 1 s will be :-

→$H = h_2 - h_1$

→$H = 19.6- 4.9$

→$H = 14.7m$