Number 55. Thanks! A 0.2-kg ball is thrown upward with an initial speed of 24 m/

Question

Number 55. Thanks!

A 0.2-kg ball is thrown upward with an initial speed of 24 m/s at an angle of 30 degree with the horizontal from the edge of a 60-m cliff. Neglecting air resistance, find (a) the initial kinetic energy of the ball, (b) the potential energy and kinetic energy of the hall when it is at its highest point, (c) the kinetic energy of the ball when it reaches the ground 60 m below the cliff, and (d) the speed of the ball just before it strikes the ground. Assume that U = 0 at the top of the cliff. A small block is attached to a rubberlike material that exerts a force F_x = -kx - a x^2 when it is stretched a distance x(x > 0), where k and a are constants. Find the work on the block by the material when it is stretched

Explanation / Answer

vx = v cos theta = 24 cos 30 = 20.78 m/s

vy = v sintheta = 24 sin30 = 12 m/s

v = sqrt vx^2 + vy^2 = sqrt ( 20.78)^2 + ( 12)^2 = 24 m/s

(a)

KEi = 1/2 m v^2 = 1/2 ( 0.2) ( 24)^2 = 57.6 m/s

(b)

h = u^2 sin^2 theta/ 2g = ( 24)^2 sin ^2 30/ 2( 9.8) = 7.34 m

PE = mg(H+h) = 0.2 ( 9.8) ( 7.34) = 14.38 J

at height point vertical velocity is zero only it ball has horizontal velocity

KE = 1/2 m (vx)^2 = 1/2 ( 0.2) ( 20.78)^2 = 43.18 m/s

( c)

y = yi + vyt + 1/2 ay t^2

-60 m = 0+ (12) t - 1/2 ( 9.8) t^2

4.9 t^2-12 t - 60 m = 0

solving t = 4.93 s

vyf = vy + ay t

= 12 + (-9.8) ( 4.93 s) = -36.31

vf = sqrt vxf^2 + vyf^2 = sqrt ( 20.78)^2 +( -36.31)^2 = 41.83 m/s

(c)

KEf = 1/2 m vf^2 = 1/2 ( 0.2) ( 41.83)^2 = 175 m/s

(d)

speed of the ball when strikes the ground is

vf = 41.83 m/s

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