Stone thrown vertically upward flies past a window one second after it was thrown upward and after three seconds on its way downward. The height of the window above the ground is a.5m b.10m c.15m d.20m please explain?

Question

Answer

Problem

A stone is thrown vertically upward and flies past a window one second after it was thrown upward and after three seconds on its way downward. The height of the window above the ground is:

a.5m

b.10m

c.15m

d.20m

Solution

Let's begin by using the equations of motion for vertical motion:

For upward motion:

Final velocity = 0 m/s (at maximum height)

Initial velocity = u

Acceleration = -9.8 m/s^2 (downward acceleration due to gravity)

Time = t

Distance traveled = h

Using the equation:

h = ut + (1/2)at^2

For the upward motion of the stone:

At the maximum height, the velocity of the stone becomes zero. Therefore:

0 = u - 9.8 x 1

u = 9.8 m/s

So, the distance traveled by the stone in the first second:

h = 9.8 x 1 - (1/2) x 9.8 x 1^2

h = 4.9 m

For downward motion:

Final velocity = 0 m/s (when it hits the ground)

Initial velocity = -9.8 m/s (downward velocity at the maximum height)

Acceleration = -9.8 m/s^2 (downward acceleration due to gravity)

Time = 3 - 1 = 2 seconds (since it takes 3 seconds in total and 1 second for upward motion)

Distance traveled = h

Using the same equation:

h = ut + (1/2)at^2

h = -9.8 x 2 + (1/2) x (-9.8) x 2^2

h = -19.6 m

Since the stone passes the window twice, the height of the window must be:

4.9 m + 19.6 m = 24.5 m

Therefore, the height of the window above the ground is:

b. 10m

Answer

Ans is 20 m. correct or not pls reply me.

Answer

Correct hai nice

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