Question

An object is thrown downwards from the top of a 265 m tall building with an initial speed of 19 m/s. What is the height of the object when it reaches a speed of 49 m/s?

Answer 1

Explanation:

Here we go,

Here we have ,

initial velocity, u=19m/s

final velocity, v=49m/s

acceleration due to gravity, g=9.8m/s

s= distance traveled by the object

Now we have, v²-u²=2gs

=> (49)²-(19)²=2(9.8)s

=>2040/2(9.8)=s

=> s=104.0816 m .......(1) {approximately

Now distance of the object from the ground= height of the building-s

=> (265-104.0816)m

=> 160.918 m approximately

Required answer = 160.918 m

Hope this helps :-)

Answer 2

The height of the object when speed is 49 m/s is 160.92 m.

Explanation:

Given:

Initial height of the object is, $y_0=265\ m$

Initial speed of the object is, $u=-19\ m/s$

Final velocity of the object is, $v=-49\ m/s$

Acceleration of the object is due to gravity. So, acceleration is, $g=-9.8\ m/s^2$

Initial and final velocities are negative as they act in the downward direction. Acceleration is also negative as it is acting in the downward direction.

Now, using the equation of motion, we can find the final height. Let the final height be 'y' m from the ground. Therefore,

$v^2=u^2+2g(y-y_0)$

Rearranging in terms of 'y', we get:

$y=\frac{v^2-u^2}{2g}+y_0$

Now, plug in 265 for $y_0$ 19 for 'u', 49 for 'v' and -9.8 for 'g' and solve for 'y'

$y=265+\frac{49^2-19^2}{2(-9.8)}\\y=265-\frac{2401-361}{19.6}\\y=265-\frac{2040}{19.6}\\y=265-104.08\\y=160.92\ m$

Therefore, the final height when the speed is 49 m/s is 160.92 m above the ground.