Question

A ball is thrown vertically upwards .It goes to a height of 45m and then returns to the ground .Taking g as 10ms^-2. Find:

i) the initial velocity of the ball
ii) the final velocity of the ball on reaching the ground
iii) the total time of journey of the ball​

Answer 1

${\huge{\boxed{\mathcal{\red{Answer}}}}}$

i) 30m/s

ii) 30m/s

iii) 6s

${\huge{\boxed{\mathcal{\red{Explanation}}}}}$

i) It is given that :

Maximum height reached by ball = 45m

Acceleration due to gravity = 10m/s^2

Initial velocity = u = ?

Now at maximum height the final velocity of an object is 0 , thus for this case

Final velocity = 0m/s

Now, according to third equation of motion :

${v}^{2} = {u}^{2} + 2as$

$s \: = \frac{ {v}^{2} - {u}^{2} }{2as}$

After inputting the values in this equation we get :

$45 = \frac{ { - u}^{2} }{2 \times - 10}$

$45 \times 2 \times 10 = {u}^{2}$

$900 = {u}^{2}$

$u = \sqrt{900} = 30$

Thus the initial velocity of ball in this case is 30m/s.

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ii) Now if we talk about acceleration of body it is given as 10m/s^2 which is equal to :

$10 = \frac{v - u}{t}$

$10 = \frac{ - 30}{t}$

$t = 30 \div 10 = 3s$

Now the time that it takes to get on maximum height is 3s.

Now it's final velocity on touching the ground will be : (Here final velocity in first case will be initial velocity.)

According to first equation of motion :

$v = u + at$

$v = 0 + 10 \times 3$

$v = 30$

Thus final velocity of ball is 30m/s.

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iii) Now, since values of (g) remains constant thus time taken to reach maximum height = time taken to reach ground.

Time taken by ball to reach max. height = 3s

Thus time taken to come down = 3s

Now total time taken in journey of ball = 3+3 = 6s.

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BY√\____________ SCIVIBHANSHU

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