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

Time taken to go to the top = 8/2 s = 4 s

Acceleration = g = - 10 m/s² (Upward)

(i) Final velocity = 0 m/s, g = 10 m/s².

We know,

$v_f = v_i \pm at$

or $v_f = v_i \pm gt$

⇒ $v_i = gt$

⇒ $v_i = (10 \ ms^{-2})(4 \ s) = 40 \ ms^{-1}$

(ii) We know,

$2a \Delta x = {v_f}^2 - {v_i}^2$

or, $2g \Delta h = {v_i}^2$

⇒ $\Delta h = \dfrac{{v_i}^2}{2g}$

⇒ $\Delta h = \dfrac{(40 \ ms^{-1})^2}{2 \times 20 \ ms^{-2}} = 80 \ m$

(iii) We know,

$\Delta x = v_i t \pm \frac{1}{2} at^2$

or, $\Delta h = v_i t \pm \frac{1}{2} gt^2$

⇒ $\Delta h = (40 \ ms^{-1})(2 \ s) - \frac{1}{2} (10 \ ms^{-2})(2 \ s)^2$

⇒ $\Delta h = 60 \ m$.

∴ Initial velocity of the ball was of 40 m/s, maximum height attained by it is 80 m and distance travelled by the ball after 2 seconds of motion is 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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