- #1
LotusTK
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There is a question i have done in my textbook and it initially asks you to calculate:
the maximum height of the ball, and how long it takes to arrive at that maximum height.
They give you a vertical velocity component of 5m/s. So i made an equation using t= (v-u)/a
Which gives me t=(0-5)/-9.81 so time = 0.51 seconds.
And with this time i calculated the maximum height using s=ut+1/2 at^2
So i did s = 5 x 0.51 + 0.5 x -9.81 x 0.51^2 = 1.274m
The next part of the question says that a tennis court has a length of 24 metres with a net of height 1.1m, and it asked if the ball would clear this net.
So i deduced that the net will be 12 metres away from the point of launch at the ball. So i did a calculation of d=vt in order to find what horizontal distance the ball had traveled when it is at its maximum height.
d= 20 x 0.51 (the horizontal velocity was given in the question, and the time will remain the same)
so d= 10.2 metres.
This is short of the 12 metres. HOWEVER. I then did another calculation of d=vt, but i changed the d to 12 metres and i kept the velocity the same. I did this in order to find what height the ball would be at, when it traveled the horizontal distance of 12 metres. (how high is it when it arrives at the net)
so t =12/20 = 0.6 seconds.
I then plugged this 0.6 seconds back into the earlier equation of s=ut +1/2 at^2 i used.
s = 5 x 0.6 + 0.5 x -9.81 x 0.6^2 = 1.2342m
Comparing this calculation to other shows that the ball will begin to fall at 10.2 metres, but at 12 metres (at the net), it is still at a height of 1.2342 metres, which is high enough to clear the 1.1m net.
So i concluded that eventhough the ball is moving downwards on its parabolic motion when at arrives at the net, it will still clear the net, as it will be 0.1342 metres above the net.
Long explanation but i think i have said it clearly enough? So is all of the above correct? Have i made any errors?
I had to ask here as there are no answers in the back of the book and there are none online... :(
Thanks
the maximum height of the ball, and how long it takes to arrive at that maximum height.
They give you a vertical velocity component of 5m/s. So i made an equation using t= (v-u)/a
Which gives me t=(0-5)/-9.81 so time = 0.51 seconds.
And with this time i calculated the maximum height using s=ut+1/2 at^2
So i did s = 5 x 0.51 + 0.5 x -9.81 x 0.51^2 = 1.274m
The next part of the question says that a tennis court has a length of 24 metres with a net of height 1.1m, and it asked if the ball would clear this net.
So i deduced that the net will be 12 metres away from the point of launch at the ball. So i did a calculation of d=vt in order to find what horizontal distance the ball had traveled when it is at its maximum height.
d= 20 x 0.51 (the horizontal velocity was given in the question, and the time will remain the same)
so d= 10.2 metres.
This is short of the 12 metres. HOWEVER. I then did another calculation of d=vt, but i changed the d to 12 metres and i kept the velocity the same. I did this in order to find what height the ball would be at, when it traveled the horizontal distance of 12 metres. (how high is it when it arrives at the net)
so t =12/20 = 0.6 seconds.
I then plugged this 0.6 seconds back into the earlier equation of s=ut +1/2 at^2 i used.
s = 5 x 0.6 + 0.5 x -9.81 x 0.6^2 = 1.2342m
Comparing this calculation to other shows that the ball will begin to fall at 10.2 metres, but at 12 metres (at the net), it is still at a height of 1.2342 metres, which is high enough to clear the 1.1m net.
So i concluded that eventhough the ball is moving downwards on its parabolic motion when at arrives at the net, it will still clear the net, as it will be 0.1342 metres above the net.
Long explanation but i think i have said it clearly enough? So is all of the above correct? Have i made any errors?
I had to ask here as there are no answers in the back of the book and there are none online... :(
Thanks
