Question

Please help me!!!
How much force is required to launch a 1 kg pumpkin 10 meters away from 5 meters off the
ground at a launch angle of zero degrees if it takes the catapult 0.1 seconds to fully launch the
pumpkin?

Answer 1

Answer:

F= 99 N

Explanation:

Dynamics and Kinematics

This problem is a combination of dynamics and kinematics because we need formulas and concepts of both branches of physics to solve it.

We need to calculate the force required to launch horizontally a pumpkin of m=1 Kg a distance of d=10 m away from a height of h=5 m.

Since the force is:

F = m.a

We need to calculate the acceleration required to move the pumpkin from rest (vo=0) to the launching speed (vf) in a time t=0.1 seconds.

The acceleration can be calculated by using the kinematic equation:

$\displaystyle a=\frac{v_f-v_o}{t}$

The final launching speed vf can be calculated knowing the height and maximum horizontal distance reached by the pumpkin.

When an object is thrown horizontally with a speed vf from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v_f\cdot\sqrt{\frac {2h}{g}}$

Solving for vf:

$\displaystyle v_f=d\cdot\sqrt{\frac {g}{2h}}$

Substituting:

$\displaystyle v_f=10\cdot\sqrt{\frac {9.8}{2\cdot 5}}$

Calculating:

$v_f=9.9\ m/s$

Now we calculate the acceleration:

$\displaystyle a=\frac{9.9-0}{0.1}$

$a= 99\ m/s^2$

Thus, the force required is:

$F=1\ Kg\cdot 99\ m/s^2$

F= 99 N