Question

A body is thrown horizontally with the speed of 10m/s from the top of the building having height 20m,then the time it will take to reach the ground while neglecting air friction is

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Time\:taken=0.82\:sec}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a body is thrown horizontally with the speed of 10m/s from the top of the building having height 20m.

• We have to find the time it will take to reach ground.

$\underline \bold{Given : } \\ \implies Initial \: velocity(u) = 10 \: m/s \\ \\ \implies Height(s) = 20 \: m \\ \\ \implies Acceleration(a) = 10 \: m /{s}^{2} \\ \\ \underline \bold{To \:Find : } \\ \implies Time \: taken = ?$

• According to given question :

$\bold{Using \: second \: equation \: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {v}^{2} = {20}^{2} + 2 \times 10 \times 20 \\ \\ \implies {v}^{2} = 400 + 400 \\ \\ \implies {v}= \sqrt{800} \\ \\ \bold {\implies v = 20 \sqrt{2} \: m/s} \\ \\ \bold{Using \: first \: equation \: of \: motion : } \\ \implies v = u + at \\ \\ \implies 20 \sqrt{2} = 20 + 10 \times t \\ \\ \implies 20 \sqrt{2} - 20 = 10t \\ \\ \implies t = \frac{20( \sqrt{2} - 1) }{10} \\ \\ \implies t = 2 \sqrt{2} - 2 \\ \\ \implies t = 2.82 - 2 \\ \\ \bold{\implies t = 0.82 \: sec}$

Answer 2

Solution:

Given:

➜ A body is thrown horizontally with the speed of 10m/s from the top of the building having height 20m.

Find:

➜ Time taken.

According to the given question:

➜ 10 m/s = initial velocity.

➜ 20 m = height.

➜ 10 m/s² = Acceleation.

Using second equation of motion:

➜ V² = u² + 2 as

➜ V² = 20² + 2 × 10 × 20

➜ V² = 400 + 400

➜ V = √800

➜ V = 20 √2m/s

Using first equation of motion:

➜ V = u + at

➜ 20 √2 = 20 + 10 × t

➜ 20 √2 - 20 = 10t

➜ t = 20 (√2 - 1)/10

➜ t = 2 √2 - 2

➜ t = 2.82 - 2

➜ t = 0.82 sec.

Therefore, the time taken is 0.82 seconds.

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