NEEDS HELP PLEASE!!
A 100kg shot putter is putting a shot 4kg. Just before he releases the shot, he is in the air (feet are off the ground). At this instant, the only force he exerts against the shot is an 800N horizontal force directed forward.
a-what is his horizontal acceleration at this instant?
b-what is his vertical acceleration at this instant?
c-what is the shots horizontal acceleration at this instant?
d-what is the shots vertical acceleration at this instant?
To find the answers to these questions, we need to apply the principles of Newton's second law of motion (F = ma) and consider the forces acting on the shot putter and the shot.
a) Horizontal acceleration of the shot putter:
The only horizontal force acting on the shot putter is the 800N force directed forward. Since we have the force and the mass of the shot putter, we can use Newton's second law to find the acceleration:
F = ma
800N = (100kg + 4kg) * a (mass of shot putter + mass of the shot)
804N = 104kg * a
a = 804N / 104kg
a ≈ 7.73 m/s²
b) Vertical acceleration of the shot putter:
Since there is no vertical force acting on the shot putter, the vertical acceleration at this instant is approximately 0 m/s².
c) Horizontal acceleration of the shot:
By Newton's third law of motion, the shot exerts the same amount of force on the shot putter, but in the opposite direction. Therefore, the horizontal acceleration of the shot is the same as the shot putter's acceleration:
a_shot = 7.73 m/s²
d) Vertical acceleration of the shot:
Since the shot is in the air and there are no vertical forces acting on it, the vertical acceleration of the shot is approximately 0 m/s².
Remember to consider that these calculations are made at the instant just before releasing the shot and with the assumption of no other external forces acting on the system.