An object is thrown vertically upward with a speed of 19.1 m/s. How high does it rise?
at the top, the velocity is zero.
vf^2 =vi^2 + 2 g h
g=-9.8m/s^2 solve for h.
345.4
To determine how high the object rises, we can use the kinematic equation for vertical motion:
\(v_f^2 = v_i^2 + 2a\Delta y\)
Where:
\(v_f\) = final velocity (which is 0 m/s at the maximum height)
\(v_i\) = initial velocity (the speed at which the object is thrown upward)
\(a\) = acceleration due to gravity (-9.8 m/s\(^2\))
\(\Delta y\) = change in height (which is what we want to find)
Rearranging the equation to solve for \(\Delta y\):
\(v_f^2 - v_i^2 = 2a\Delta y \)
Since the object comes to rest at its maximum height, the final velocity (\(v_f\)) is 0 m/s. Plugging in the values we have:
\(0^2 - 19.1^2 = 2(-9.8)\Delta y \)
Simplifying the equation:
\(0 - 365.81 = -19.6\Delta y \)
\(365.81 = 19.6\Delta y \)
To find \(\Delta y\), we divide both sides of the equation by 19.6:
\(\Delta y = \frac{365.81}{19.6} \)
Simplifying:
\(\Delta y \approx 18.67\) meters
Therefore, the object rises to a height of approximately 18.67 meters.