A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 130 m/s from the top of a cliff 150 m above level ground, where the ground is taken to be

y = 0.

(a) What is the initial total mechanical energy of the projectile? (Give your answer to at least three significant figures.)
answer = 535680 J

(b) Suppose the projectile is traveling 92.1 m/s at its maximum height of y = 326 m. How much work has been done on the projectile by air friction?
________ J

(c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?

_______ m/s

Explanation / Answer

(a) Here, intially projectile is at 150 m above the ground and its speed is 130 m/s

So, the total mechanical energy of the projectile = Ke+Pe

=1/2mv^2+mgh

=54*(1/2x130^2+9.8x150)

= 535680 J

(b) Now, at the top of its flight its vertical velocity will be zero,it will have only horizontal velocity.

Again, there is no friction, so horizontal velocity will remain constant.

And its magnitude = 130*cos30 = 112.6 m/s, but due to friction its velocity is reduced to 92.1 m/s

Again, its intial vertical veolocity = 130*sin30 = 65 m/s

It is given body has reached upto 326 m from ground,as it wasthrown from cliff which is 150 m above ground distance travelled byprojectile vertically from the point of throw = 326 - 150 =176 m

we shall calculate the real vertical speed due tofriction using y=176 m

Now use the formula -

v^2-u^2=2gs;

at the top V=0

so U^2=2x9.8 m/s^2x176

=> U = 58.7 m/s

So, due to friction new velocity of projectile is 92.1i+ 58.7j

Magnitude= (92.1^2 + 58.7^2) = 109.2 m/s

So, work done by air =1/2m Vi^2- 1/2 mVf^2

=1/2x54(130^2 - 109.2^2) = 1.34 x 10^5 J

(c) Work done by air in upward movemnt = 1.34 x 10^5 J

It is givenair has done 1.5 time more work downward;

So, total work done by air = 2.5x1.34 x 10^5

Initial ke of body was=1/2x54x130^2

final ke= 1/2x54xV^2

final ke= intial ke - work done by friction+mgx150 as projectile reaches ground

=> 1/2x54V^2=1/2x54x130^2 - 2.5x1.34 x 10^5 + 54x10x150

=> 27*V^2 = 456300 - 335000 + 81000

=> V = 86.6 m/s.

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