Question

When you drop a 4.5 kg watermelon off the Dennison building, it reaches a terminal velocity of around 35 m/s second before smashing into the ground. If the collision between the watermelon and the ground lasts 3 milliseconds, what is an estimate of the average force the watermelon exerts on the ground while it is landing?

1. 7600 N
2. 420 N
3. 52500 N
4. 106000 N

Answer 1

Answer:

3) 52500N

Explanation:

We use the mathematical definition of impulse as follows;

$Ft = mv............(1)$

were F is the force experienced, t is the time of impact, m is the mass of the object and v is the velocity of impact.

Given; m = 4.5kg, v = 35m/s, t = 3 milliseconds

however, $3milliseconds=3*10^{-3}s$

From equation (1);

$F=\frac{mv}{t}.............(2)$.

Substituting all necessary values into equation (2), we obtain the following;

$F=\frac{4.5*35}{3*10^{-3}}}\\\\F= \frac{157.5.5}{0.003}\\\\F=52500N$