Question

g A rock is thrown downward from the top of a 38.4-m-tall tower with an initial speed of 16 m/s. Assuming negligible air resistance, what is the speed of the rock just before hitting the ground

Answer 1

Final velocity, V = 31.76m/s

Step-by-step explanation:

Given the following data;

Initial velocity = 16m/s

Distance = 38.4m

We know that acceleration due to gravity is equal to 9.8m/s².

To find the final velocity, we would use the third equation of motion;

`V^(2) = U^(2) + 2aS`

Where;

• V represents the final velocity measured in meter per seconds.

• U represents the initial velocity measured in meter per seconds.

• a represents acceleration measured in meters per seconds square.

• S represents the displacement measured in meters.

Substituting into the equation, we have;

`V^(2) = 16^(2) + 2*9.8*38.4`

`V^(2) = 256 + 752.64`

`V^(2) = 1008.64`

Taking the square root, we have;

`V^(2) = \sqrt {1008.64}`

Final velocity, V = 31.76m/s