Question is to long so i put it in explanation box PLEASE HELP!
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 m/s at an angle of 24o above the horizontal.
a) How long does it take for the ball to reach the wall if it is 4.2 m away? - s
b) How high is the ball when it hits the wall
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a) You should remember the equation that relates the distance, velocity and time such that:
`D = v*t`
The problem provides the velocity and the total distance, hence, you should evaluate the time such that:
`t = D/v`
You need to consider the horizontal component of velocity such that:
`v_x = v cos 24^o`
`t = D/(v*cos 24^o) => t = 4.3/(14*cos 24^o) => t = (4.3 m)/(12.789 m/s)`
`t = 0.33 ` s
Hence, evaluating the time taken for the ball to hit the wall yields `t = 0.33 s` .
b) You know that the ball hits the wall after `t = 0.33 s,` hence, using the vertical component of velocity, you may evaluate the height reached by the ball when it hits the wall such that:
`h = v_y*t => h = 14*sin 24^o*0.33`
`h ~~ 1.879 m`
Hence, evaluating how high is the ball when it hits the wall yields `h ~~ 1.879 m` .
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