Download the projectile motion physics simulation from the PhET website: http://

Question

Download the projectile motion physics simulation from the PhET website: http://phet.colorado.edu/en/simulation/projectile-motion

C. Select a different type of projectile with a different mass. Keep your angle and firing velocity the same, and fire the new projectile. How does the trajectory of the new projectile compare to that of the old one? Is this a surprising result? Explain your answer by referencing the basic equations of motion with constant acceleration and/or Newton’s laws.

D. Go back to your original type of projectile. Keeping everything else constant, what angle will allow your projectile to travel the farthest horizontal distance? What angle will allow your projectile to reach the greatest vertical height?

E. Set your angle to some value between 10o and 30o and fire the cannon. Move the bull’s-eye to where your projectile landed. Without changing anything else, can you find another firing angle that is significantly larger than 30o but less than 90o that will launch your projectile to the same spot (ie, dead center of the bull’s-eye)? What were your two angles? Find another pair of angles that will launch your projectile to the same location; what is the relationship between these pairs of angles? Keep in mind that the simulation has an uncertainty of about +/- 2 degrees; this will obscure the relationship

Explanation / Answer

(c) ANSWER:

the projectile motion does not dependends upon weight of the body

in the projectle motion some values to be constant

like acceleration is constant

the acceletation value is 9.8 in earth

the basic equations of projectile motion is

maximum height =u^2 SIN^2(theta) /2g

horizantal distance R=U^2 sin2(theta)/g

so the projectile motion is does not depends upon mass of body

(D) ANSWER:

(1)for given value of projection velocity u.R is horizantal distance is maximum

when 2(theta) =90

(theta) =45

so maximum horizantal distance is

Rmax =u^2/g {sin(90)=1}

(2) when (theta) =90 Hmax =u^2/2g

this is equal to the maximum height reached by a body projected vertically upwards

the angle of maximum horizantal distance is = 45

the angle of maximum vertical distance is =90

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