Two-Dimensional Kinematics 434

Question

In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14.9 m/s at an angle of 23.3° above the horizontal. How long does it take for the ball to reach the wall if it is 4.06 m away? How high is the ball when it hits the wall?

Answer 1

S=Vi*t*cos(23.3)...you got S and Vi so get (t)..then

h=Vi*t*sin(23.3) - (g/2)*(t^2)....g=9.8 m/s^2....get h

Answer 2

My first step is to determine the time for the ball to reach its maximum height.

The equation is given by

Vf^ = Vo(sin 23.3) - gT

where

Vf = final velocity = 0 (when ball reaches its peak)

Vo = initial velocity = 14.9 m/sec.

g =  acceleration due to gravity = 9.8 m/sec^2 (constant)

T = time to reach maximum height

Substituting appropriate values,

0 = 14.9(sin 23.3) - (9.8)T

Solving for T,

T = 0.60 sec.

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To reach a wall 4.06 m away,

X = Vo(cos 23.3)(T) = 4.06

T  = 4.06/(14.9 * cos 23.3)

T = 0.297 sec.  --- this simply shows that the ball will hit the wall 4.06 meters away before it reaches its maximum height.

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The height of the ball when it hits the wall is given by the equation

Y = Vo(cos 23.3)(T) - (1/2)(gT^2)

where

Y = vertical height along the wall where the ball will hit

and all the other terms have been previously defined.

Substituting appropriate values,

Y = (14.9)(sin 23.3)(0.297) - (1/2)(9.8)(0.297)^2

Y = 1.32 meters (above ground level)