A basketball is thrown at 45° to the horizontal. The hoop is located 4.5m away horizontally at a height of 1m above the point of release. What is the required initial speed?
Horizontal
Vcos45*t=4.5
Vertical:
Vsin45*t-4.9t^2=1
solve for t in the first equation (in terms of V), put that into the second, and solve for V
Not good explanation
how do u solve v?
Put t=4.5/Vcos45 into the second equation, you only have V as the unknown. It is a quadratic, use the quadratic equation.
To find the required initial speed of the basketball, we can use the equations of motion for projectile motion.
First, we need to break down the motion into horizontal and vertical components. The initial velocity can be represented as two vectors:
Vx (horizontal component) = V * cos(θ)
Vy (vertical component) = V * sin(θ)
Given that the hoop is located 4.5m away horizontally and 1m above the point of release, we can determine the time it takes for the basketball to reach the hoop using the vertical motion equation:
h = h0 + Vy0 * t + (1/2) * g * t^2
Since the height at the point of release (h0) is 0, and the height at the hoop (h) is 1m, we can solve for the time (t).
1 = 0 + (V * sin(θ)) * t + (1/2) * (-9.8) * t^2
1 = (V * sin(θ)) * t - 4.9 * t^2
Similarly, we can determine the horizontal distance traveled by the basketball using the horizontal motion equation:
x = x0 + Vx0 * t
Since the initial horizontal position (x0) is 0, and the final horizontal position (x) is 4.5m, we can solve for the time (t).
4.5 = 0 + (V * cos(θ)) * t
4.5 = (V * cos(θ)) * t
Now, we have two equations with two unknowns (V and t). We can solve these equations simultaneously to find the required initial speed (V).
Dividing the second equation by the first equation, we get:
4.5 / 1 = (V * cos(θ)) * t / ((V * sin(θ)) * t - 4.9 * t^2)
Simplifying:
4.5 = V * cos(θ) / (V * sin(θ) - 4.9 * t)
Next, we substitute the value of t from the second equation into the above equation:
4.5 = V * cos(θ) / (V * sin(θ) - 4.9 * (4.5 / (V * cos(θ))))
Simplifying further:
4.5 = V * cos(θ)^2 / (V * sin(θ) - 4.9 * (4.5 / V))
Multiply both sides by (V * sin(θ) - 4.9 * (4.5 / V)):
4.5 * (V * sin(θ) - 4.9 * (4.5 / V)) = V * cos(θ)^2
Expand:
4.5 * V * sin(θ) - 4.9 * (4.5 / V) * V = V * cos(θ)^2
Simplify:
4.5 * V * sin(θ) - 4.9 * 4.5 = V * cos(θ)^2
Rearrange the equation:
V * cos(θ)^2 - 4.5 * V * sin(θ) + 4.9 * 4.5 = 0
Now, we have a quadratic equation in terms of V (the initial speed). We can solve this equation using the quadratic formula:
V = [-b ± √(b^2 - 4ac)] / (2a)
where:
a = cos(θ)^2
b = -4.5 * sin(θ)
c = 4.9 * 4.5
Substituting the values of a, b, and c into the quadratic formula, we can find the two possible values for V.