Question

a flowerpot dropped from a window and fell for 3.3 seconds to the ground. How high was the window?

Answer 1

s=ut+1/2 at^2

s=0+1/2 9.8(3.3)^2

s=53.34m

Answer 2

Concept:

• One - dimensional motion
• Kinematic equations of one-dimensional motion
• Free-fall of an object

Given:

• Flowerpot is dropped, so it has no initial velocity
• Time taken for the flowerpot to fall to the ground = 3.3 s
• Value of acceleration due to gravity = 9.8 m/s^2

Find:

• The height at which the window was

Solution:

We know the formula for one-dimensional kinematics.

s = ut +1/2gt^2

where s is the displacement, u is the initial velocity, g is the acceleration due to gravity and t is the duration of the fall.

u = 0 m/s, g = 9.8 m/s^2, t = 3.3 s

s = 0+1/2(9.8) *(3.3)^2

s = 53.4 m

The window was 53.4 m above the ground.

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