Respuesta :
Velocity v0 = 13.00m/s
Height h = 2.16 m
At the maximum height V = 0,
a) θ = 45 degrees
Calculating the velocity while reaching the peak
Vf = v0sinθ = 13 x sin45 = 9.19 m/s
Calculating the time taken, V = Vf + gt => Vf - gt = 0 => t = Vf / g
=> t = 9.19 / 9.81 => t = 0.937s
time taken for the descent will be same t = 0.937s
Total Time T = 0.937 + 0.937 = 1.874s
Now the horizontal velocity Vh = v0cosθ = 13 x cos45 = 9.19 m/s
horizontal distance s = Vh x T = 9.19 x 1.874 = 17.22 meters
b) θ = 42 degrees
Calculating the velocity while reaching the peak
Vf = v0sinθ = 13 x sin42 = 8.67 m/s
Calculating the time taken, V = Vf + gt => Vf - gt = 0 => t = Vf / g
=> t = 8.67 / 9.81 => t = 0.883s
time taken for the descent will be same t = 0.883s
Total Time T = 0.883 + 0.883 = 1.767s
Now the horizontal velocity Vh = v0cosθ = 13 x cos42 = 9.66 m/s
horizontal distance s = Vh x T = 9.66 x 1.767 = 17.07 meters
a) s = 17.22 m
b) s = 17.07 m
Height h = 2.16 m
At the maximum height V = 0,
a) θ = 45 degrees
Calculating the velocity while reaching the peak
Vf = v0sinθ = 13 x sin45 = 9.19 m/s
Calculating the time taken, V = Vf + gt => Vf - gt = 0 => t = Vf / g
=> t = 9.19 / 9.81 => t = 0.937s
time taken for the descent will be same t = 0.937s
Total Time T = 0.937 + 0.937 = 1.874s
Now the horizontal velocity Vh = v0cosθ = 13 x cos45 = 9.19 m/s
horizontal distance s = Vh x T = 9.19 x 1.874 = 17.22 meters
b) θ = 42 degrees
Calculating the velocity while reaching the peak
Vf = v0sinθ = 13 x sin42 = 8.67 m/s
Calculating the time taken, V = Vf + gt => Vf - gt = 0 => t = Vf / g
=> t = 8.67 / 9.81 => t = 0.883s
time taken for the descent will be same t = 0.883s
Total Time T = 0.883 + 0.883 = 1.767s
Now the horizontal velocity Vh = v0cosθ = 13 x cos42 = 9.66 m/s
horizontal distance s = Vh x T = 9.66 x 1.767 = 17.07 meters
a) s = 17.22 m
b) s = 17.07 m
Effectively, a shot putter at the given configuration which shoots a ball at an angle of 42º will have a horizontal reach higher than if he shoots at an angle of 45º. If he shoots at an angle of 42º the ball will reach 19.23 meters, and if he shoots at 45º the ball will reach 19.21 meters.
Further explanation
Our problem is the following:
Given that an object (in this case a ball) is shot with a speed of 13 meters per second, at an angle [tex]\theta[/tex] above the horizontal, from a point which is 2.16 meters above the ground... Calculate the horizontal distance at which the object touches the ground.
This is a free fall 2-dimensional type of problem. To solve it, we must model how the ball moves over time, for this we apply the equations of kinematics for free falling objects and obtain that:
[tex]x= V \cdot cos(\theta) \cdot t[/tex]
[tex]y= y_o + V \cdot sin(\theta) \cdot t - \frac{g \cdot t^2}{2}[/tex]
Where [tex]V[/tex] is the initial speed of the object (in this case, 13 m/s), [tex]y_0[/tex] is the initial height of the ball (2.16 meters for this problem), and [tex]g[/tex] is the gravity (whose value is [tex]9.8 \frac{m}{s^2}[/tex]).
We know that at a given time, let's call it [tex]t_a[/tex], the ball will touch the ground (and y will be zero at that time). We wish to compute [tex]t_a[/tex] so that we can also compute the ball's horizontal reach. From the second equation we can then write that:
[tex]0= y_0 + V \cdot sin(\theta) \cdot t_a - \frac{g \cdot {t_a}^2}{2}[/tex]
From the above we can solve for [tex]t_a[/tex] by using the quadratic formula, therefor:
[tex]t_a = \frac{V\cdot sin(\theta) + \sqrt{(V \cdot sin(\theta))^2 + 4\cdot \frac{g}{2}\cdot y_0}}{2\cdot \frac{g}{2}}[/tex]
Plugging numbers we find that, for 45º [tex]t_a[/tex] is 2.09 seconds, and for 42º [tex]t_a[/tex] is 1.99 seconds.
Then plugging [tex]t_a[/tex] on the equation for the horizontal movement of the ball, we find that for the 45º shot, the ball travels 19.21 meters:
[tex]x_{45 \º}= 13 \cdot cos(45 \º) \cdot 2.09 = 19.21[/tex]
And for the 42º shot the ball travels 19.23 meters:
[tex]x_{42 \º}= 13 \cdot cos(42 \º) \cdot 1.99 = 19.23[/tex]
Learn more
Here is a list of problems similar to yours, they're about parabolic motion:
- https://brainly.com/question/10932946
- https://brainly.com/question/659054
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Keywords
Parabolic motion, free fall, gravity
