Question

Joseph stands on top of a building 120 meters high. He throws a ball vertical downward with an initial velocity of 25 m/s. The acceleration due to gravity is 9.8 m/s^2. (a)What is the velocity of the ball after it falls for 2 seconds?& &
(b)What is the position of the ball after 2 seconds?
(c)what time did the ball reach the ground?
(d)What is the velocity of the ball as it strikes the ground?

Answer 1

Joseph stands on top of a building 120 meters high. He throws a ball vertical downward with an initial velocity of $25\,\mathrm{m/s}$. The acceleration due to gravity is $9.8\,\mathrm{m/s^2}$.

(a) What is the velocity of the ball after it falls for 2 seconds?

(b) What is the position of the ball after 2 seconds?

(c) What time did the ball reach the ground?

(d) What is the velocity of the ball as it strikes the ground?

Solution

If the height of building is $H = 120\,\mathrm{m}$, initial velocity is $v_{in} = 25\,\mathrm{m/s}$, acceleration due to gravity is $g = 9.8\,\frac{\mathrm{m}}{\mathrm{s}^2}$ we have

(a) The velocity of ball is

$$\begin{array}{l} v(t) = v_{in} + gt \\ t = 2\,\mathrm{s} \\ v = 19.6\,\mathrm{m/s} + 25\,\mathrm{m/s} = 44.6\,\mathrm{m/s} \end{array}$$

(b) The ball is situated at a height of

$$\begin{array}{l} h = H - \frac{g t^2}{2} - v_{in} t \\ t = 2\,\mathrm{s} \\ h = 120\,\mathrm{m} - 19.6\,\mathrm{m} - 50\,\mathrm{m} = 50.4\,\mathrm{m} \end{array}$$

(c) We have when ball strikes the ground:

$$\begin{array}{l} h_f = 0 \\ 0 = H - \frac{g t^2}{2} - v_{in} t \Rightarrow \\ t = \frac{-v_{in} + \sqrt{v_{in}^2 + 2Hg}}{g} \approx 3,016 \end{array}$$

(d) According to the law of conservation of energy we have (velocity of the ball is $v_f$)

$$\begin{array}{l} \frac{m v_{in}^2}{2} + mg H = \frac{m C^2}{2} + m g h \\ h_f = 0 \\ v_f = \sqrt{v_{in}^2 + 2mg H} \approx 54,56\,\mathrm{m/s} \end{array}$$