hf=hi+vivertical*time-4.9 t^2
vivertical=14.9*sin55
solve for time in air.
vivertical=14.9*sin55
solve for time in air.
H = H1 + V1y*t - (1/2)*g*t^2,
where:
H is the final height (H2 = 14.0 m),
H1 is the initial height (43.0 m),
V1y is the initial vertical component of the velocity (V1 * sin(θ)),
g is the acceleration due to gravity (9.81 m/s^2), and
t is the time.
First, let's calculate V1y:
V1y = V1 * sin(θ)
= 14.9 m/s * sin(55.0°).
Next, we can rearrange the equation to solve for time:
H2 = H1 + V1y*t - (1/2)*g*t^2
14.0 m = 43.0 m + (14.9 m/s * sin(55.0°)) * t - (1/2)*9.81 m/s^2 * t^2.
This equation is quadratic in t. We can rearrange it to the form: at^2 + bt + c = 0:
(1/2)*9.81 m/s^2 * t^2 - (14.9 m/s * sin(55.0°)) * t - (43.0 m - 14.0 m) = 0.
Now, we can solve this quadratic equation using the quadratic formula:
t = [-b ± √(b^2 - 4ac)] / (2a),
where a = (1/2)*9.81 m/s^2, b = -(14.9 m/s * sin(55.0°)), and c = -(43.0 m - 14.0 m).
Plugging in the values, we can solve for t:
t = [-(14.9 m/s * sin(55.0°)) ± √((14.9 m/s * sin(55.0°))^2 - 4*(1/2)*9.81 m/s^2 * (43.0 m - 14.0 m))] / (2*(1/2)*9.81 m/s^2).
Calculating this expression will give us two values for t. However, we are only interested in the positive value since time cannot be negative in this context. Thus, we disregard the negative value.
Finally, the positive value for t will give us the time the ball is in the air.