Question

a ball thrown up vertically returns to the thrower after six seconds. find the velocity with which it was thrown up ,the maximum hight it reaches, and the position after four seconds ​

Answer 1

Question :

A ball thrown up vertically returns to the thrower after six seconds. find the velocity with which it was thrown up ,the maximum hight it reaches, and the position after four seconds .

Solution :

FreeFall

• Motion of an object under acceleration due to gravity (g) is termed as Freefall

• g varies with height

• Freefall is symmetrical if air resistance is neglected

• Acceleration due to gravity is 10 m/s²

From the Question,

• Total Time Taken,T = 6 s

• Acceleration,g = 10 m/s²

Since,

The ball returns to the thrower,the displacement and final velocity of the ball are zero

Using the Relation,

$\boxed{ \boxed{ \sf{v = u - gt}}}$

The ball is thrown up, according to the sign convention,g = - 10 m/s²

Substituting the values,we get :

$\longmapsto \sf{ - u = - 10 \times 6} \\ \\ \longmapsto \: \boxed{ \boxed{ \sf{u = 60 \: m {s}^{ - 1} }}}$

Maximum Height

Using the Relation,

$\boxed{ \boxed{ \sf{h = ut - \frac{1}{2} g {t}^{2} }}}$

Remember that freefall is symmetrical,thus the ball would take 3s to reach the top and another 3s to reach the ground again » t = 3s

$\longmapsto \: \sf{h = (60)(3) - \dfrac{1}{2} \times 10 \times 3 \times 3 } \\ \\ \longmapsto \: \sf{h = 180 - 45} \\ \\ \longmapsto \: \boxed{ \boxed{ \sf{h = 135 \: m}}}$

When t = 4s,

$\longmapsto \: \sf{h = (60)(4) - \dfrac{1}{2} \times 10 \times {4}^{2} } \\ \\ \longmapsto \: \sf{h = 240 - 80} \\ \\ \longmapsto \: \boxed{ \boxed{ \sf{h = 160 \: m}}}$

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.