If we assume that the initial velocity of the projectile is v, its launch angle is θ, and the acceleration due to gravity is g, then the height y of the projectile at a time t seconds after launch is given by the following quadratic equation:
y = (v^2 * sin(2θ)) / (2g) - g*t^2/2
This equation has two variables, y and t, and two constants, v and g. We can solve for y using the quadratic formula:
Answers & Comments
Answer:
Step-by-step explanation:
If we assume that the initial velocity of the projectile is v, its launch angle is θ, and the acceleration due to gravity is g, then the height y of the projectile at a time t seconds after launch is given by the following quadratic equation:
y = (v^2 * sin(2θ)) / (2g) - g*t^2/2
This equation has two variables, y and t, and two constants, v and g. We can solve for y using the quadratic formula:
y = (-(v^2 * sin(2θ)) / (2g)) + square root(v^2 * sin(2θ)^2 / 4g^2 + v^2 * cos(2*theta))
We can then substitute v^2 * sin(2θ) / (2g) with -p, v^2 * cos(2*theta) with q, and v^2 with p^2 + q^2 to get:
y = (-p/2 + sqrt(p^2 + q^2)) + square root(p - sqrt(p^2 + q^2))
Now we have an equation in the form of the formula y = -p/2 + sqrt(p^2 + q^2) + sqrt(p^2 + q^2 - a), where a is a constant equal to -p*q/2.
To find the time at which the projectile reaches its maximum height, we can differentiate the equation with respect to t and set equal to zero:
dy/dt = -g
Solving for t, we get:
t = -2sqrt(v^2 * sin(2theta)) / g
Assuming v > 0, sin(2*theta) = 1, and g = -9.81 (m/s^2), we get:
t = sqrt(2v^2 / -9.81) = sqrt(2v^2 / 9.81)
So, assuming an initial velocity of 35 m/s