Hello Guest!
You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at the point even with the roof railing with an upward speed of 15.0 m/s, the ball is then in free fall. On its way back down, it just misses the railing. At the location of the building, g = 9.8m/s^2.
Find :
(a) the position and the velocity of the ball 1.0s and 4.0s after leaving your hand.
(b) The velocity when the ball is 5.0m above the railing.
(c) The maximum height reached and the time at which it is reached.
(d) The acceleration of the ball when it is at its maximum height.
(e) Find the time when the ball is 5.0m below the roof railing.
(a) the position and the velocity of the ball 1.0s and 4.0s after leaving your hand.
\(h=vt-\frac{1}{2}gt^2\\ h_{1.0}=\frac{15m}{s}\cdot 1s-\frac{1}{2}\cdot\frac{9.8m}{s^2}\cdot 1^2s^2\\ \color{blue}h_{1.0}=10.1m\ over\ the\ railing\)
\(h_{4.0}=\frac{15m}{s}\cdot 4s-\frac{1}{2}\cdot\frac{9.8m}{s^2}\cdot 4^2s^2\\ h_{4.0}=60m-79.2m=-19.2m\\ \color{blue}h_{4.0}=19.2m\ under\ the\ railing\)
\(v=v_0-gt\\ v_{1.0}=\frac{15m}{s}-\frac{9.8m\cdot1s}{s^2}=5.2\frac{m}{s}\\ \color{blue}v_{1.0}=5.2\frac{m}{s}\ up \)
\(v_{4.0}=\frac{15m}{s}-\frac{9.8m\cdot4s}{s^2}=-24.2\frac{m}{s}\\ \color{blue}v_{4.0}=24.2\frac{m}{s}\ down\)
(b) The velocity when the ball is 5.0m above the railing.
\(v=v_0-\sqrt{2gh}\\ v_{(+5m)}=\frac{15m}{s}-\sqrt{\frac{2\cdot 9.8m\cdot 5m}{s^2}}=5.1\frac{m}{s}\\ \color{blue}v_{(+5m)}=5.1\frac{m}{s}\ up\)
(c) The maximum height reached and the time at which it is reached.
\(E=\frac{mv^2}{2}=mgh\\ h=\frac{mv^2}{2mg}=\frac{v^2}{2g}\)
\(h_{max}=\frac{15^2m^2s^2}{2\cdot 9.8ms^2}\\ \color{blue} h_{max}=11.480m\ over\ the\ railing. \)
\(v=gt\\ t_{rise}=\frac{v}{g}=\frac{15ms^2}{9.8ms}\\ \color{blue} t_{rise}=1.531s\)
(d) The acceleration of the ball when it is at its maximum height.
The ball is weightless throughout the journey.
For the ball, the acceleration is zero everywhere. It is not exposed to any force.
For the viewer the acceleration is at the highest point and in the case g = 9.8m / s²
(e) Find the time when the ball is 5.0m below the roof railing.
\(s=vt-\frac{1}{2}gt^2\\ \frac{1}{2}gt^2-vt+s=0\\ t^2-\frac{2v}{g}t+\frac{2s}{g}=0\)
\(t=-\frac{p}{2}+\sqrt{(\frac{p}{2})^2-q}\\ \color{blue} t=\frac{v}{g}+\sqrt{\frac{v^2}{g^2}-\frac{2s}{g}}\)
\(t=\frac{15m/s}{9.8m/s^2}+\sqrt{(\frac{15}{9.8}s)^2-\frac{-2\cdot 5m}{9.8m/s^2}}\\ t=1.531s+\sqrt{2.343s^2+1,020s^2}\\ t=1.531s+1.834s\\ \color{blue}t=3.365s\ (5m\ under\ the\ railing)\)
!
