Question

ball is thrown vertically upwards with a velocity of 25m/s. If g is 10m/s, then calculate:

a. Height it reaches

b. Time taken to return back
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Answer 1

hope this answer will help you

Answer 2

Given: initial velocity = 25m/s

           acceleration due to gravity g = 10m/s

To Find: height = h

              time taken to return = t = t₁ + t₂

Solution:

Formulas Used:

the three equations of laws of motion are used, given as follows:

1. First Law of motion equation ⇒ v = u + gt
2. Second Law of motion equation ⇒ s/h = ut + 1/2 gt²
3. Third Law of motion equation⇒   v² = u² + 2gh

The situation mentioned in the question can be divided into two cases:

Case I: as the ball goes up, reaching the highest point, the velocity becomes zero.

Hence final velocity- v = 0

           initial velocity = 25m/s (Given)

                                g = -10m/s²(as the ball travels upwards,against the                        .                                         gravitational force.

let time taken for this journey = t₁

Case II: When the ball falls down from its highest point back to the ground.

             initial velocity u' = 0m/s

let time taken for this journey =  t₂

Case I: finding t₁ and h

Applying equation 3

v² - u² = 2gh

0 - 25 × 25 = 2 × -10 × h

- 625 = -20 × h

h = 625/20

h  =  31.25m              

Applying equation 1.

v = u + gt₁

0 = 25 - 10t₁

-25 = -10t₁

t₁ = -25/-10

t₁ = 2.5 seconds

Case II : finding t₂

Applying equation 2.

h = ut + 1/2gt₂²

31.25 = 0 + 1/2 × 10t₂²

31.25 = 5t₂²

t₂² = 31.25/5

t₂² = 6.25

t₂ = $\sqrt{6.25}$

t₂ = 2.5 seconds

t = t₁ + t₂

t = 2.5 + 2.5

t = 5 seconds

∴ a) height the ball reaches - h= 31.25m

   b)Time taken to return back t = 5 seconds