A boy standing on top of a building throws a small ball from a height of H1 = 43.0 m. The ball leaves with a speed of 14.9 m/s, at an angle of 55.0 degrees from the horizontal, and lands on a building with a height of H2 = 14.0 m. Calculate for how long the ball is in the air. (Neglect air friction, and use g = 9.81 m/s2.)
hf=hi+vivertical*time-4.9 t^2
vivertical=14.9*sin55
solve for time in air.
To calculate the time the ball is in the air, we can use the equation of motion for vertical motion:
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.