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

Answer:

Final velocity, V = 31.76m/s

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