Answer to Question #206459 in Physics for Morgan Hopkin

Question

Miss Takken throws a 10 g ball upwards at an initial speed of 4.5 m/s while standing on a 37 m high cliff. 

 

a.     What maximum height will the ball reach above the ground?

b.     What will the ball’s height be when it is traveling at 5.6 m/s?

c.     What will the ball’s potential energy be just before impact?

d.     What will the ball’s speed be just before impact?

Answer 1

(a) Let's first find the time the ball takes to reach maximum height:

$$v=v_0-gt,$$ $$0=v_0-gt,$$ $$t=\dfrac{v_0}{g}=\dfrac{4.5\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=0.46\ s.$$

Then, we can find the maximum height the ball reach above the ground from the kinematic equation:

$$y_{max}=y_0+v_0t-\dfrac{1}{2}gt^2,$$ $$y_{max}=37\ m+4.5\ \dfrac{m}{s}\cdot0.46\ s-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(0.46\ s)^2,$$ $$y_{max}=38.03\ m.$$

(b) Let's first find the time the ball takes to reach the velocity of 5.6 m/s after it passes its maximum height:

$$v=v_0-gt,$$ $$v=-gt,$$ $$t=\dfrac{v}{-g}=\dfrac{-5.6\ \dfrac{m}{s}}{-9.8\ \dfrac{m}{s^2}}=0.57\ s.$$

Then, we can find the height when the ball is traveling at 5.6 m/s from the kinematic equation:

$$y=y_{max}-v_0t-\dfrac{1}{2}gt^2,$$ $$y=38.03\ m-5.6\ \dfrac{m}{s}\cdot0.57\ s-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(0.57\ s)^2,$$ $$y=33.24\ m.$$

(c) According to the law of conservation of energy, just before the impact all the potential energy will be converted to kinetic energy. Therefore, the potential energy just before the impact equals zero:

$$PE_f=0.$$

(d)

$$PE_i+KE_i=PE_f+KE_f,$$ $$PE_i=KE_f,$$ $$mgy_{max}=\dfrac{1}{2}mv^2,$$ $$v_f=\sqrt{2gy_{max}}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot38.03\ m}=27.3\ \dfrac{m}{s}.$$