Is My Kinematic Solution for a Basketball Shot Correct?

[Chazz569]

I'm working on this problem and I've came up with a solution but however I couldn't find anything else to verify the problem so if you could let me know if I did it correctly and help me on the right track what would be super.

The problem goes as folllow: A player is shooting a ball into a net, he shooting with a 45 degree angle and is 4m away from the net and the net is one meter up. How fast must he throw the ball for it to go in (initial speed).

I used the equation: D=v1*t+1/2*a*t^2 and split it into X and Y which gives me these equations:
Y: 1=v1*t+1/2*(-9.8)*t^2 (we allways assume -9.8m/s^2 for gravity)
X: 4=v1*t+1/2* (0) *t^2
and solved it gived me 5.11 m/s but since that the X/Y speed we want the longest line of the triangle so I do
Cos(45)=5.11208/x
x=7.22957m/s
So my answer would be 7.22957m/s

Is this correct or I am way off?

 

[whozum]

$$x(t) = V_0cos(45)t$$
$$y(t) = V_0sin(45)t + \frac{gt^2}{2}$$

You want x(t) to be 4
and y(t) to be 1

You should get a system of equations which you solve for V_0. Your answer seems a bit high.

 

[marlon]

The second formula should be :

$$y(t) = V_0sin(45)t - \frac{gt^2}{2}$$

because the y-axis is pointing upwards and gravity is downwards

marlon

 

[whozum]

What I would probably do is calculate your time of flight. You are traveling 4m right and 1m up.

$$T_{flight} = sqrt{\frac{2h}{g}}$$ from the equation you provided.

Then take the time of flight, divide the distance traveled in the x direction by it and you'll get $$V_x$$

 

[Chazz569]

Hey thanks for you help. I used the equations and I got 7.22957m/s as well so I'm guessing it's safe to assume I got the right answer? Thanks for your help, it's really nice to have resources like this forums to get some help :)

edit: I just saw your last reply whozum, give me time to check with your equations as well and I'll get back to you guy with the answer I get.

edit2: Sorry but what does the h stand for in that equation? I'm guessing the G stand for gravity which in my case would be -9.8m/s

 

[whozum]

h is the height to travel, h=1m

For V in the x direction

$$T = sqrt{2h/g} = sqrt{2/9.8} = 0.452$$

$$d = 4m, t = 0.452s, v = d/t$$

$$v_x = \frac{4m}{0.452s} = 8.85m/s$$

For V in the y direction

$$y(t) = V_ysin(45)t+\frac{gt^2}{2}$$

Solving for V_y:

$$V_y = \frac{1-\frac{gt^2}{2}}{sin(45)t}$$

$$V_y = \frac{1-\frac{(-9.8)(0.452)^2}{2}}{sin(45)(0.452)}$$

$$V^2 = V_x^2+V_y^2$$

 

[Chazz569]

Well that's gives me the Sqrt of -.204082 which is a non-real number.

 

[whozum]

Take g to be positive 9.8 instead of -9.8. Look at my work above.