Question

b) A ball is thrown up and caught back by the thrower after 8 second. Calculate (i) the velocity with which the ball was thrown up, (ii) the maximum height attained by the ball and (iii) distance covered by the ball after 2 second of the start of motion. (Take g = 10 m/s2

Answer 1

$\dag\:\underline{\sf AnsWer :}$

In the question it is given that, a ball is thrown up and caught back by the thrower after 8 second. Hence, time taken by the ball to reach maximum height is 8/2 = 4 second, final velocity (v) will be 0 m/s [beacuse ball will return back to the thrower] and Acceleration due to gravity (g) is -10 m/s² [As the ball is thrown upward the value of g will be negative]. and we are asked to find (i) the velocity with which the ball was thrown up [Initial Velocity (u) = ?] (ii) the maximum height attained by the ball = ? and (iii) distance covered by the ball after 2 second of the start of motion = ?

• First we will find the velocity with which the ball was thrown up by using first kinematical equation of motion :

$:\implies\sf v = u + gt$

$:\implies\sf 0= u + ( - 10) \times 4$

$:\implies \underline{ \boxed{\sf u = 40 \: m/s}}$

• We have find the the velocity with which the ball was thrown up i.e u = 40 m/s. Now, we can find the the maximum height attained by the ball by using second kinematical equation of motion :

$\dashrightarrow\:\:\sf s = ut + \dfrac{1}{2} gt^2$

$\dashrightarrow\:\:\sf s = 40 \times 4 + \dfrac{1}{2} \times ( - 10) \times (4)^2$

$\dashrightarrow\:\:\sf s = 160 - 5 \times 16$

$\dashrightarrow\:\:\sf s = 160 - 80$

$\dashrightarrow\:\: \underline{ \boxed{\sf s = 80 \: m}}$

• Hence,the maximum height attained by the ball is 80 m. Now, let's find the distance covered by the ball after 2 second of the start of motion by using second kinematical equation of motion :

$\longrightarrow\:\:\sf s = ut + \dfrac{1}{2} gt^2$

$\longrightarrow\:\:\sf s = 40 \times 2 + \dfrac{1}{2} \times ( - 10) \times (2)^2$

$\longrightarrow\:\:\sf s = 80 - 5\times 4$

$\longrightarrow\:\:\sf s = 80 - 20$

$\longrightarrow\:\: \underline{ \boxed{\sf s = 60 \: m}}$

Answer 2

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Given:-

• final velocity (v) = 0 m/s
• acceleration due to gravity = - 10 m/s² (because the ball is thrown upwards or against the gravity)
• total time = 8 seconds

Time taken to reach the required height:-

$\longrightarrow$ time = $\sf\dfrac{8}{2}$ = 4 seconds

To find:-

• the velocity with which the ball was thrown i.e the initial velocity

formula to be used:-

$\underline{\boxed{\sf v = u+gt}}$

here,

• v = final velocity
• u = initial velocity
• g = acceleration due to gravity
• t = time taken

$\longrightarrow$ v = u+gt

$\longrightarrow$ 0 = u+ -10×4

$\longrightarrow$ 0 = u + -40

$\longrightarrow$ u = 40 m/s

hence, the initial velocity is 40 m/s

_____________________________

To find:-

• maximum height attained by the ball

formula to be used:-

$\underline{\boxed{\sf v²-u² = 2gh}}$

here,

• v = final velocity
• u = initial velocity
• g = acceleration due to gravity
• h = height attained

Solution:-

$\longrightarrow$ 0²-40² = 2 × (-10) × h

$\longrightarrow$ -1600 = -20 × h

$\longrightarrow$ h = $\sf\cancel\dfrac{-1600}{-20}$ = 80 m

hence, the required height is 80 m

______________________________

To find:-

• distance covered by the ball in 2 seconds

Formula to be used:-

$\underline{\boxed{\sf S = ut + \dfrac{1}{2} gt ²}}$

here,

• S = distance
• u = initial velocity
• t = time taken
• g = acceleration due to gravity

Solution:-

$\longrightarrow$ S = 40×2 + $\sf\dfrac{1}{2}$ × (-10) × 2 × 2

$\longrightarrow$ S = 80 + (-20)

$\longrightarrow$ S = 80-20

$\longrightarrow$ S = 60 m

hence, the distance travelled by the ball after 2 seconds is 60 m.

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