How long does it take for the ball to reach the wall if it is 4.2 m away and what is the height at which the ball hits the wall?
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 24 degrees above the horizontal.
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The ball is struck at ground level and sent towards a wall with a speed of 14 m/s at an angle of elevation 24 degrees above the horizontal.
The initial velocity of the ball can be divided into two components, one parallel to the horizontal equal to 14*cos 24 and the other perpendicular to the horizontal equal to 14*sin 24 in the upward direction.
As the wall is 4.2 m away, the time taken by the ball to reach it is `4.2/(14*cos 24) ~~ 0.3284` seconds.
To determine the height at which the ball strikes the wall the acceleration due to gravity that is 9.8 m/s^2 in the vertically downwards direction has to be kept in mind. The height of the ball is equal to `14*sin 24*0.3284 - (1/2)*9.8*(0.3284)^2`
`~~ 1.3415 m`
The balls takes approximately 0.3284 seconds to reach the wall and strikes it at a height 1.3415 m.
I found the time but it is saying that the height is wrong. Please help!
a) You need to remember the equation that puts in relation the distance, velocity and time such that:
`d = vt`
Since the problem provides the value of velocity and the total distance, hence, you need to substitute these values in equation above such that:
`t = d/v`
Considering the horizontal component of velocity yields:
`v_x = v cos 24^o`
`t = 4.2/(14*cos 24^o) => t = (4.2m)/(12.789 m/s) => t = 0.32 ``s`
Hence, evaluating the time taken for the ball to reach the wall yields `t = 0.32 s.`
b) You know that the ball hits the wall after `t = 0.32 ` 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*0.32`
`h ~~ 1.82 m`
Hence, evaluating how high is the ball when it hits the wall yields `h ~~ 1.82 m.`
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