In Fig. 4-11, a ball is launched with a velocity of 7.50 m/s, at an angle of 50° to the horizontal. The launch point is at the base of a ramp of horizontal length d1 = 4.00 m and height d2 = 3.60 m. A plateau is located at the top of the ramp.
What is the magnitude of its displacement from the launch point when it lands?
I just mainly want to make sure my math is right since this is my last try.
vy=7.5cos50
vy=5.745
dymax=vo^2/2g
dymax=(5.745)^2/19.6
dymax=1.684
My one question here is whether or not I assume this is where it hits the ramp, since there is no way to reach the plateau. If not, how would I find where it actually hits the ramp.
dymax=1/2gt^2
t=(2dymax/g)^1/2
t=.586
dx=vxt
dx=4.821 x .586
dx=2.825
I didn't use the time for it to go back down since the ball hits the ramp. Where am I going wrong?
Well, it seems like you're on the right track, but let me add some humor to lighten things up and help clarify your question:
Oh, the physics of launching balls and ramps! It's enough to make you feel like you're rolling downhill without brakes! But fear not, I'm here to bring some chuckles and, hopefully, answers to your questions.
First off, assuming that you've calculated the vertical component of the initial velocity (vy) correctly as 5.745 m/s, and you've found the maximum height (dymax) to be 1.684 meters, good job! These calculations seem to be spot-on.
Now, as for your question about where the ball hits the ramp, it's essential to consider that the plateau is at the top of the ramp, not the landing point. So, the ball will hit the ramp somewhere before reaching the plateau. Determining this exact spot requires a bit more analysis.
To find the horizontal displacement (dx) when the ball lands, you'll need to calculate the time it takes for the ball to hit the ramp after being launched. But worry not, my friend; we can't let gravity go unchallenged! Let's do some more calculations.
Your equation for finding the time for the ball to reach its maximum height is correct: t = (2dymax/g)^1/2, which gives you t ≈ 0.586 seconds. Applause for your mathematical prowess!
But wait, there's more! We need to consider the time it takes for the ball to reach the ramp from its maximum height. To find this time (let's call it tramp), you can use the equation: tramp = (2d2/g)^1/2 since d2 is the vertical distance between the maximum height and the ramp. So, tramp = (2*3.60/9.8)^1/2 ≈ 0.851 seconds. Ta-da!
Finally, we can find the total time (ttotal) for the ball's flight. We add the time it took to reach the ramp (tramp) to the time it took to reach the maximum height (t): ttotal = t + tramp ≈ 0.586 + 0.851 ≈ 1.437 seconds. Time sure flies when you're doing math!
Now, multiplying the total time (ttotal) by the horizontal component of the initial velocity (4.00 m/s) will give you the horizontal displacement (dx): dx = 1.437 * 4.00 ≈ 5.748 meters.
So, my friend, the magnitude of the displacement from the launch point when the ball lands is approximately 5.748 meters. Hurray!
Remember, humor always helps in understanding the ups and downs of physics. Keep up the good work, and if you have any more questions or need further calculations, I'm here to clown around and assist you!
To find the magnitude of the displacement from the launch point when it lands, you need to break down the problem into two parts: the horizontal displacement and the vertical displacement.
Let's start with the vertical displacement. You correctly determined the initial vertical velocity of the ball to be vy = 7.5 cos(50°) = 5.745 m/s. Using this value, you calculated the maximum height reached by the ball, which is correct: dymax = (5.745)^2 / (2 * 9.8) = 1.684 m.
Now, to find where the ball lands, you need to consider the horizontal displacement. The time it takes for the ball to come back down from its maximum height can be found using the formula: t = sqrt((2 * dymax) / g) = sqrt((2 * 1.684) / 9.8) = 0.586 s.
Since the ball was launched horizontally, its horizontal velocity remains constant throughout its trajectory. Therefore, the horizontal displacement is given by the formula: dx = vx * t.
To find the horizontal velocity vx, you can use the initial velocity of the ball and the launch angle. vy = 7.5 sin(50°) = 5.748 m/s. This is the initial vertical velocity, and it remains constant throughout the motion. The horizontal velocity, vx, can be found using the formula: vx = vy / tan(50°) = 5.748 / tan(50°) = 5.748 / 1.1918 = 4.825 m/s.
Now, you can calculate the horizontal displacement: dx = vx * t = 4.825 * 0.586 = 2.825 m.
So, the magnitude of the displacement from the launch point when it lands is approximately 2.825 meters.
Your calculations seem correct, so it appears you were on the right track. Just make sure you carefully consider the horizontal and vertical components of the motion separately and use the correct equations for each.
To find the magnitude of the displacement from the launch point when the ball lands, you need to consider the horizontal and vertical components of the motion separately.
First, let's calculate the time it takes for the ball to reach its maximum height (dymax) using the vertical component of the velocity. The vertical component of the velocity (vy) is given by vy = vo * cos(angle), where vo is the initial velocity (7.50 m/s) and angle is the launch angle (50°).
vy = 7.50 * cos(50°)
vy ≈ 5.745 m/s
Next, find the time it takes to reach the maximum height using the equation dymax = vy^2 / (2 * g), where g is the acceleration due to gravity (-9.8 m/s^2).
dymax = (5.745)^2 / (2 * 9.8)
dymax ≈ 1.684 m
This calculation gives you the maximum height of the ball in the vertical direction.
Now, let's find the time it takes for the ball to fall from the maximum height and hit the ground. Since the launch point is at the base of a ramp, assume the ball hits the ramp.
Using the equation dymax = (1/2) * g * t^2, where t is the time it takes to fall, and solving for t:
t = sqrt((2 * dymax) / |g|)
t = sqrt((2 * 1.684) / 9.8)
t ≈ 0.5855 s
This calculation gives you the time it takes for the ball to fall from the maximum height and hit the ramp.
Finally, calculate the horizontal displacement (dx) using the horizontal component of the velocity (vx) and the time it takes for the ball to hit the ramp:
vx = vo * sin(angle)
vx = 7.50 * sin(50°)
vx ≈ 5.710 m/s
dx = vx * t
dx = 5.710 * 0.5855
dx ≈ 3.344 m
Therefore, the magnitude of the displacement from the launch point when the ball lands is approximately 3.344 m.