Question

A ball thrown vertically upward returns after 8s. Calculate
(a) the velocity with which it was thrown up.
(b) the maximum height attained.
(c) its position after 6 s.
Please Solve This​

Answer 1

Answer:

hmmmmm i do not know sorry i will try

Answer 2

Solution  :-

Total time (T) = 8 sec

We know that ,

Time of ascent = Time of descent = t

T = t + t

T = 2 t

t = 8 / 2

t = 4 s

• Time of ascent = Time of descent = t = 4 s

Ball moving upwards

• final velocity of the ball (v)= 0 m/s
• Time taken to reach maximum height(t) = 4 sec
• Acceleration due to gravity (g) = - 10 m/s ( as the ball is moving against gravity )

A the ball is moving with uniform  acceleration throughout it's motion we can use kinematic equation in order to solve such type of questions

By using first kinematic equation ,

$\longrightarrow\mathtt{ v = u+ gt }$

Now let's substitute the given values in the above equation ,

$\longrightarrow\mathtt{ 0 = u +(-10) 4 }$

$\longrightarrow\mathtt{ u = 40 \dfrac{m}{s} }$

The ball was thrown up with a velocity of 40 m /s

We can easily find the maximum height of the ball by using  second or third kinematic equation.

By using the third equation of motion ,

$\longrightarrow\mathtt{ v^{2} = u^{2}+ 2gh }$

Now let's substitute the given values in the above equation ,

$\longrightarrow\mathtt{ 0 = 1600+ 2(-10)h }$

$\longrightarrow\mathtt{ 20 h = 1600 }$

$\longrightarrow\mathtt{ h = 80 m}$

The maximum height obtained by the ball = 80 m

After 4 seconds the ball starts to descend in downward direction

We need to find the position of the ball after 6 sec i.e. 4 s(upward)+ 2s(downwards)

Ball moving in downward direction

• initial velocity (u') = 0 m/s ( velocity at highest point is zero )
• time = 2 s
• Acceleration due to gravity = 10 m/s ( as ball is moving along gravity )

By using the second equation of motion ,

$\longrightarrow\mathtt{ h' = u' t + \dfrac{1}{2}gt^{2} }$

Now let's substitute the given values in the above equation ,

$\longrightarrow\mathtt{ h' = \dfrac{1}{2}\times 10\times 4 }$

$\longrightarrow\mathtt{ h' = 20 m }$

The position of the ball from the ground after 6 sec = 80 - 20 = 60 m