Question

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate the maximum height to which it rises. (ii) The total time it takes to return to the surface of the earth​

Answer 1

Answer:

Initial velocity of the ball (u) = 49m/s.

The velocity of the ball at maximum height (v) = 0.

g = 9.8m/s2

Find out

1) The maximum height to which it rises.

2) The total time it takes to return to the surface of the earth.

Let us consider the time is t to reach the maximum height H.

Consider a formula,

2gH = v2 – u2

2 × (- 9.8) × H = 0 – (49)2

– 19.6 H = – 2401

H = 122.5 m

PLEASE MARK AS BRAINLIEST.

Answer 2

Given:

• initial velocity of ball (u) = 49 m/s
• final velocity of ball (v) = 0 m/s

To find:

• i) the maximum height at which it rises,

• ii) the total time it takes to return to the surface of the earth.

formula used:

$\bigstar\:{\underline{\boxed{\frak{\purple{v^2 - u^2 = 2gh}}}}}$

$\bigstar\:{\underline{\boxed{\frak{\red{v = u + g \times t}}}}}$

solution:

• let's consider the time taken is "t" to reach the maximum height "h".

⠀⠀⠀⠀

now, by using the 3rd eqⁿ of motion,

$: \implies\sf{v^2 - u^2 = 2gh}$

${\dag}\:{\underline{\frak{putting\:given\:values\:in\:formula,}}}$

$:\implies\sf (0)^2 - (49)^2 = 2 \times (-9.8) \times h$

$:\implies\sf 0 - 2401= - 19.6 \times h$

$:\implies\sf - 2401 = - 19.6 h$

$:\implies\sf h = \cancel{\dfrac{- 2401}{ - 19.6}}$

$:\implies{\frak{ \red{h = 160\:m}}}\:\bigstar$

again, by using 1st eqⁿ of motion,

⠀⠀⠀⠀

$: \implies\sf{v = u + g \times t}$

${\dag}\:{\underline{\frak{by\:putting\:given\:values\:in\:formula,}}}$

$:\implies\sf 0 = 49+ (-9.8) \times t$

$:\implies\sf 0 = 49- 9.8t$

$:\implies\sf 9.8t = 49$

$:\implies\sf t = \cancel{\dfrac{49}{9.8}}$

$:\implies{\frak{ \pink{t = 5\:sec\:}}}\:\bigstar$

$\therefore\:{\underline{\sf{the\:total\:taken\:to\:reach\:the\:ground\:is\: (5 + 5) = {\textsf{\textbf{10\:sec}}}.}}}$