Question

A ball is thrown vertically upwards with a velocity of 20m/s from the top of a multistorey building.The height of the point where the ball is thrown 25m from the ground.How long will it be before the ball hits the ground. (g=10m/s^2)


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Answer 1

Your Answer:

Given:-

• Initial Velocity when ball is thrown up = 20m/s
• Total height when it falls down from reaching at the maximum height = 45
• Height of Building = 25m

To find:-

• The time taken by ball to reach at the ground

Solution:-

We know that

$\tt v = u +gt \rightarrow\rightarrow\rightarrow(1)$

from eq(1) we can find the time taken by the ball to reach at maximum distance.

So, we have

v = 0

u = 20

taking g = 10 in appx

$\tt 0 = 20+(-10)t \\\\ \tt \Rightarrow 10t = 20 \\\\ \tt \Rightarrow t = \dfrac{20}{10} \\\\ \tt \Rightarrow t = 2s$

Now, we have to find the time taken by the ball to reach at the ground from the maximum distance

we have,

s = 45

ut = 0

g = 10

So, we can use this formula

$\tt s = ut + \dfrac{1}{2}gt^2$

So, replacing values

$\tt \Rightarrow 45 = 0+ (\dfrac{1}{2}\times 10T^2) \\\\ \tt \Rightarrow 45 = 5T^2 \\\\ \tt \Rightarrow \dfrac{45}{5} = T^2 \\\\ \tt \Rightarrow 9 = T^2 \\\\ \tt \Rightarrow \pm 3 = T$

So, Time = 3s, as time can not be in negative

So, Total Time = t + T = 2s + 3s = 5s

Answer 2

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