Hi I am trying to make a computer program that calculates when an object hits the ground (on a 2d plane) given a certain angle (Degrees), speed (m/s), and an optional offset for x and or y. Is there a single equation i can use to solve this?
Example Equation:
X(t)=20TCos(30)
Y(t)=8+((20TCos(30))-4.9T^2)
What is T when Y=0 given the offset of y+8??
The equation for solving without an offset i believe is:
T=(2Vsin(angle))/9.8
Kurdt
Aug25-08, 09:32 AM
Your equation for the y distance is wrong. There should be a sine n there. If you want to find t you will have to solve the quadratic equation of the y distance. The time you have given is one solution to the quadratic equation when c is zero.
Hi I am trying to make a computer program that calculates when an object hits the ground (on a 2d plane) given a certain angle (Degrees), speed (m/s), and an optional offset for x and or y. Is there a single equation i can use to solve this?
Example Equation:
X(t)=20TCos(30)
Y(t)=8+((20TCos(30))-4.9T^2)
What is T when Y=0 given the offset of y+8??
The equation for solving without an offset i believe is:
T=(2Vsin(angle))/9.8
As Kurdt said, one of those "cosines" should be "sine"; if the angle is the initial angle with the horizontal then the Y(t) term should involve sin(30) which is 1/2. cos(30)= sqrt(3)/2.
When Y= 0, 0= 8+ (20sin(30)T- 4.9T^2 which is the same as the quadratic equation
4.9T^2- 20sin(30)T- 8= 0. You can use the quadratic formula to solve that:
T= \frac{20sin(30)\pm\sqrt{(20sin(30))^2+ 4(4.9)(8)}}{2(4.9)}