<span>Answer:
1st, identify the givens and the unknown - this will give you parameter of what concept and formula are you going to use.
Given: m= 1200kg v initial = 95km/hr v final = 0
2nd, focus on the units - in most cases units speak for the concept
the unit of the unknown is kcal, thus its the unit of energy or work
so, W = ?
3rd, provide the appropriate formula - give formula or equation that the given and the unknown are present
since W = delta K.E =delta P.E
W= 0.5m( vf^2 - vi^2) ---> best formula
4th, Substitute the given to the formula
since 1 Joule = 1Nm 1N = 1kgms^-2 1cal = 4.19 J
we express first 95 km/hr to m/s
95km/hr x 1000m/1km x 1hr/3600sec = 26.39 m/sec
W= 0.5(1200kg)[(0^2- (26.39m/sec)^2]
W=600 kg(0 - 696.43m^2/s^2)
W=600kg(-696.43m^2/s^2)
W=417859.3Nm or 417859.3 J
W = 417859.3 J x 1 cal /4.19 J
W = 99,727.7 cal or 99.728 kcal</span>
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Explanation:
As we know that catapult is projected with speed 19.9 m/s
so here we have


similarly we have


Part a)
Horizontal displacement in 1.03 s



Part b)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part c)
Horizontal displacement in 1.71 s



Part d)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part e)
Horizontal displacement in 5.44 s



Part f)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Circularity system........…….......
F=dP/dt. So you want the momentum to change as slowly as possible in time to minimize the force. So as you catch the egg, let your hand move backward with it for awhile, slowly bringing it to a stop. If you hold your hand steady when you catch it the force due to the impact could break it.