Acrazed physics student goes to the roof of the school 14.2 meters above the ground and drops a pumpkin straight down with an initial velocity of 0 m/s how long does it take for the pumpkin to hit the ground

Time taken ≈ 1.7015 seconds

Explanation:

To answer this question, we will use the following equation of motion:

S = ut + 0.5at²

where:

S is the distance traveled = 14.2 m

u is the initial velocity = 0 m/sec

t is the time taken to reach the ground that we need to find

a is the acceleration due to gravity which is 9.81 m/sec²

Substitute with the givens to get the value of t as follows:

S = ut + 0.5at²

14.2 = 0(t) + 0.5(9.81)(t²)

14.2 = 4.905t²

t² = = 2.895

t = ±√2.895

either t = +√2.895 ≈ 1.7015 seconds .............> accepted

or t = -√2.895 ≈ -1.7015 seconds .............> rejected as time can't be negative

Hope this helps :)

the independent variable, or the one that is intentionally manipulated is the speed.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be