Question

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate:

a. the max height to which it rises.

b. total time it takes to return to surface of the earth.​

Answer 1

Answer:

Given,

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

Initial velocity of the ball (u) = 49m/s.The velocity of the ball at maximum height (v) = 0.

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

The velocity of the ball at maximum height (v) = 0.g = 9.8m/s^2

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

Consider a formula,

2gH = v^2 - u^2

2 × (- 9.8) × H = 0 – (49)^2

– 19.6 H = – 2401

– 19.6 H = – 2401H = 122.5 m

b . Now consider a formula,

v = u + g × t

0 = 49 + (- 9.8) × t

0 = 49 + (- 9.8) × t– 49 = – 9.8t

0 = 49 + (- 9.8) × t– 49 = – 9.8tt = 5 sec

0 = 49 + (- 9.8) × t– 49 = – 9.8tt = 5 sec

(1) The maximum height to ball rises = 122.5 m

(2) The total time ball takes to return to the surface of the earth = 5 + 5 = 10 sec.

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Answer 2

Answer:

$\bigstar{\bold{Maximum\:height=122.5\:m}}$

$\bigstar{\bold{Total\:time=10\:s}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Initial velocity of ball (u) = 49 m/s
• Final velocity of ball (v) = 0 m/s

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Maximum height it attains =  Distance travelled (s)
• Time taken (t)

$\Large{\underline{\underline{\bf{Solution:}}}}$

a. Maximum height it attains:

→ By the third equation of motion we know that,

  v² - u² =  2as

→ Here acceleration on the body is acceleration due to gravity which acts in the direction opposite to the motion of the body.

→ Hence a = -g = -9.8 m/s²

→ Substitute the given datas in the above equation,

   0² - 49² = 2 × - 9.8 × s

   s = -2401/-19.6

   s = 122.5 m

→ Hence the maximum height attained by the body = 122.5 m

$\boxed{\bold{Maximum\:height=122.5\:m}}$

b.Total time taken to return back

→ By the first equation of motion, we know that

 v = u + at

  where a = -g = -9.8 m/s²

→ Substituting the datas given, we get the value of time.

   0 = 49 + -9.8 × t

   -9.8 × t = -49

    t = 49/9.8

    t = 5 s

→ Since the object has to travel up and down, the total time taken would be 5 + 5 =  10 s

→ Hence the total time taken for the ball to return back = 10 s

$\boxed{\bold{Total\:time=10\:s}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² =  2as