A ball player catches a ball 3.3 seconds after throwing it vertically upward. With what speed did he throw it? What height did it reach? What is its velocity at the highest point?
Please answer and show SOLUTIONS. I'm just a fourth year high school student....I know less.....Thanks!
2007-10-01 00:46:00 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
KINDLY ANSWER!!!!!!!!!!!!!
2007-10-01 00:52:50 · update #1
All objects thrown up are pulled down by earth with an acceleration of 9.8m/s^2.
That is the speed of the ball decreases by 9.8m/s in every second.
If the initial speed is 9.8m/s then in one second its speed will decrease to zero.
If the initial speed is 2x 9.8 = 19.6 m/s then in two second its speed will decrease to zero and so on.
If the initial speed is t x 9.8 m/s, then in t second its speed will decrease to zero.
Thus the equation to find the time in which the speed becomes zero is
u = gt.
Now we can find the distance it has gone up by the following way.
Since the speed varies uniformly from u to zero in t seconds,
We find the average speed as u / 2 and the distance is calculatedas
average speed x time = [u/2] t
After reaching the highest point, i.e., after attaining
zero speed the object returns to earth.
Now the speed increases by 9.8 m/s in every second. Hence it takes the same time to reach the ground and same increase in speed and the same distance.
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Now the answer to your question is ' the highest point implies its speed is zero'. If its speed were not zero at the highest point then it will move still up and hence the point is not the highest.
Since he catches the ball after 3.3 s, the time to go up is 3.3/2 = 1.65 s
Using the equation for time and initial speed
u = gt we get u = 9.8 x 1.65 = 16.17 m/s
The distance moved up = (16.17 /2) x 1.65 = 13. 34025 m.
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2007-10-01 02:31:55 · answer #1 · answered by Pearlsawme 7 · 0⤊ 1⤋
Okay
t(up)=t(down)
Does that help? More?
V(on the top)=V0 - gt=0 (at the highest point V=0)
So V0=gt= 9.81 x (3.3/2)=16.2 m/s
h=0.5 g t^2
h=0.5 x 9.81 x (3.3/2)^2=13.4 m
Again the velocity at the highest point we know is zero like in V=0
2007-10-01 07:52:19 · answer #2 · answered by Edward 7 · 0⤊ 0⤋
A free falling object falls at 32ftps squared
2007-10-01 07:55:58 · answer #3 · answered by coach 4 · 0⤊ 1⤋
time of ascent =v/g
total time of flight =2v/g =3.3
v=3.3g/2=16.1865m/sec
ht = v*v/2g =13.35m/sec
vel at ht point =0 m/sec
because after heighest point ball will return back
2007-10-01 07:55:00 · answer #4 · answered by PRADEEP K 1 · 1⤊ 0⤋