Question

A ball is thrown up vertically returns to the thrower after 6s.Find a) the velocity with which it was thrown up, b) the maximum height it reaches, and c) it’s position after 4s

Answer 1

Given :

A ball is thrown up vertically and it gets back to the thrower after 6s.

To Find :

a)The velocity with which it was thrown up

b)The maximum height it reaches

c) It’s position after 4s

Solution :

a)The ball takes total 6s for its upward and downward journey.

We know time of ascent is equal to time of descent.

So,it would have taken 3s to attain the maximum height.

Final velocity of the ball at the maximum height,v= 0

Acceleration due to gravity,g = -9.8m/s^2

According to the first equation of motion v = u + at

Here, a = g

So, v = u + at

$\sf{}0=u+(-9.8\times3)$

$\sf{}0=u-29.4$

$\sf{}u=29.4m/s$

Hence the ball was thrown upward with a velocity of 29.4 m/s.

b)Let the maximum height attain by the ball be h.

Initial velocity,u = 0

Acceleration due to gravity,g = -9.8m/s^2

From second equation of motion,

$\sf{}s =ut + \dfrac{1}{2}at^2$

$\sf{}h=29.4\times 3+\dfrac{1}{2}\times-9.8m/s\times (3)^2$

$\sf{}h= 44.1m$

c)Ball attains the maximum height after 3s.

After attaining this height,it will start falling downward.

In this case,

Initial velocity,u = 0

Postion of the ball after 4s of the throw is given by the distance travelled by it during it’s downward fall in 1s (4s-3s)

From second equation of motion,

$\sf{}s =ut + \dfrac{1}{2}at^2$

$\sf{}s =0\times+\dfrac{1}{2}\times 9.8\times 1^2=4.9m$

Total height = 44.1m

This means that the ball is 39.2m(44.1m - 4.9m) above the ground after 4 second

Answer 2

Answer:

Explanation:

The ball returns to the ground after 6 seconds.

Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s

Let the velocity with which it is thrown up be  u

(a). For upward motion,                

v=u+at

∴     0=u+(−10)×3                  

⟹u=30  m/s      

(b). The maximum height reached by the ball

h=ut+  1 /2 at2

h=30×3+  1/2 (−10)×3  2

               

h=45 m  

(c). After 3 second, it starts to fall down.  

Let the distance by which it fall in 1 s   be   d

d=0+  1/2 at  2 ′        

where   t  =1 s

′

d=  1/2×10×(1)  2

=5 m  

∴ Its height above the ground, h  

′

=45−5=40 m  

Hence after 4 s, the ball is at a height of 40 m above the ground.