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Asst 8 Seed Question 2
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approxomation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
In order to find how high the ball rises, we first need to find the time it takes it to get to its highest point. We can use the equation ‘dt= ‘dv / a to solve for the time. When we do this, we get 15 m/s / 10 m/s/s = 1.5 s (we can assume that vf=0 m/s because the ball is still when it hits its highest point). Now that we know this number, we can solve the displacement (to get the highest point) using the equation s= s0 + vy0 * t. When we plug in the numbers we know, we can get s= 12 m + 15 m/s * 1.5 s = s= 34.5 m.
• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
Because we do not know ‘dv or ‘dt, we need to find one of these variables first. We can solve for ‘dt by using the equation y= ½ * a * ‘dt ^ 2 + v0 * ‘dt. We plug in what we know to get 34.5 m = ½ * 10 m/s/s * ‘dt ^2. We solve for ‘dt ^ 2 = 6.9 s and get the square root of this for seconds, 2.62 s. Now we can use the equation a=’dv/’dt = ‘dv= a * ‘dt = 10 m/s/s * 2.62 s = 26.2 m/s. This is the velocity final because v0 = 0 m/s.
• At what clock time(s) will the speed of the ball be 5 meters / second?
When we throw the ball upward, it has an initial velocity of 15 m/s. Simultaneously, gravity pulls at ten m/s. Therefore, after 1 s, the ball is moving at 5 m/s. We can also see this by using the equation a= (vf-v0) / (tf-t0) and solve for tf= (5 m./s – 15 m/s) / 10 m/s/s = -1 sec. This is a positive number because we are measuring time.
• At what clock time(s) will the ball be 20 meters above the ground?
We follow the same process as above (but with a different equation) to get this answer for clock time when the ball is 20 m above the ground. In this case, we use ‘dt= ‘ds / vAve = 20 m / 7.5 m/s = 2.67 sec. (We already knew that vAve was 7.5 m / s from the previous problem). To get the 2nd time the ball reaches 20 m above the ground, we see that the distance covered is 34.5 m – 20 m = 14.5 m. The vAve is ( vf + v0 ) / 2 = (26.2 m/s + 0 m/s) /2 = 13.1 m/s. The time is solved by ‘dt = ‘ds / vave = 14.5 m / 13.1 m/s = 1.107 sec.
• How high will it be at the end of the sixth second?
At the end of the 6 th second, the ball has already hit the ground. We know this by adding the time to get up (1.5 sec) plus the time to get down (2.62 sec) to get a net total time of 4.12 sec.
This is most likely correct, but there might be a hole in the ground (e.g., a wellshaft). In that case the ball would be at -90 m from its initial point, or 78 m down the hole.
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About 40 minutes.
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Very good work on this problem. See my note.
&#Let me know if you have questions. &#