Question

a crazed 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

Answer 1

To solve this problem, we use acceleration due to gravity (g), which we will assume, this close to the Earth, is 9.8 m/s/s. To find time, we look at four known kinematic equations:

d=(Vf-Vi/2)t

d=Vit + 1/2at^2

Vf^2 = Vi^2 + 2ad

Vf = Vi + at

With the given information, which is a displacement of 14.2 meters downward (-14.2), an initial velocity of 0, and acceleration due to gravity downwards (-9.8 meters per second per second), we could solve the problem by plugging in these constants for our second equation and solving for t. However, to avoid the inefficiency of possibly using the quadratic formula, we can instead solve for final velocity, and use that to find our change in time. If we look at our third equation, we can simplify it to:

Vf = Square Root Of; 2(-9.8)(-14.2)

Once you have a final velocity, solving for time becomes much easier, where

Vf = Vi + at,

or rearranged as

t = (Vf-Vi)/a

Answer 2

Time taken ≈ 1.7015 seconds

Step-by-step 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 =
`(14.2)/(4.905)` = 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 :)