Question

A 0.650 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1425 J and at its highest point is 177.6 m above the launch point. What is the horizontal component of its velocity?  

What was the vertical component of its velocity just after launch?

At one instant during its flight the vertical component of its velocity is 35.40 m/s. At that time, how far is it above or below the launch point?

Answer 1

Answer 2

SOLUTION :

Mass, m = 0.650 kg 

K.E. = 1425 J

K. E. = 1/2 m u^2  (u is initial velocity)

=> 1425 = 1/2 * 0.650 * u^2

=> u = sqrt((2*1425)/0.650)

=> u = 66.22 m/sec

Let this be at an angle xº with the horizontal/.

So, 

Horizontal component of initial velocity = u cos x = 66.22 cos x 

Vertical component of initial velocity = u sin x = 66.22 sin x 

At highest point, vertical velocity = 0 

So,  K.E is converted into P.E 

=> Velocity head = Height attained

=> (66.22 sin x)^2 / 2g = 177.6 

=> sin^2 x = 177.6 * 2 * 9.8 / 66.22^2 = 0.7938 

=> sin x = sqrt(0.7938) = 0.891

=> x = asin (0.891) = 62.9947º = 63º approx.

So,

Horizontal component of its velocity = 66.22 cos(63) = 30.0 m/sec  approx. (ANSWER)

Vertical component of its velocity = 66.22 sin (63) = 59.0 m/sec (ANSWER).

When vertical velocity is 35. 40 m/sec :

Height gained

= Initial velocity head. - Velocity head at  35.40 m/sec

= 59^2 / 2g - 35.4^2 / 2g

= (59^2 - 35.4^2) / (2*9.8)

= 113.67 m above the launch point (ANSWER).