Question

A stone is thrown vertically upward with speed of 12.5 from the edgae of ......

please show all the steps. problem #81

ch inter- 1 km/h -d limit, e lights alculate 81. A stone is thrown vertically upward with a speed of 12.5 m/s from the edge of a cliff 75.0 m high (Fig. 2-49). (a) How much later does it reach the bottom of y=0 the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel? u make her car turned 85. Bill can thro Joe can. How 86. Sketch the as a functio 87. A person do tion just as yellow light 28 m away Should she intersectior 15 m wide. whereas it Ignore the e speed e lights make it all is tance issed. g the y = -75 m FIGURE 2-49 Problem 81. 82. Figure 2-50 is a position versus time graph for the motion of an object along the x axis. Consider the time interval from A to B. (a) Is the object moving in the positive or negative direc- tion? (b) Is the object speeding up or slowing down? (c) Is the acceleration of the object positive or negative? Next, consider the time interval from D to E. (d) is the object moving in the positive or negative direction? (e) Is the object speeding up or slowing down? () is the acceleration of the object posi- tive or negative? (8) Finally, answer these same three questions for the time interval from C to D. om a green nhill. we an ange that the 88. A car is driver lo can acce cover the the rear oncomir traveling away.SE 89. Agent below, a spots a measur passing is 1.5 m many from th many CUP It is .x (m)

Answer 1

(a) How much later does it reach the bottom of the cliff?

using equation of motion (ii), we have

y = y₀ + v₀ t + (1/2) g t²

where, v₀ = initial velocity of stone = 12.5 m/s

y₀ = initial height of stone from top of the cliff = 0

y = final height of stone = - 75 m

g = acceleration due to gravity = - 9.8 m/s²

then, we get

- (75 m) = (0) + (12.5 m/s) - (0.5) (9.8 m/s²) t²

(4.9 m/s²) t² - (12.5 m/s) t - (75 m) = 0

From quadratic equation, we get

t = - b $\pm$ (_$\sqrt{}$b² - 4 a c) / 2a

t = - (-12.5 m/s) $\pm$ [_$\sqrt{}$(-12.5 m/s)² - 4 (4.9 m/s²) (-75 m)] / [(2) (4.9 m/s²)]

t $\approx$ - 2.83947 sec    OR    t = 5.39049 sec

t = - 2.84 sec           OR     t = 5.40 sec

(b) What is its speed just before hitting?

using equation of motion (i), we have

v = v₀ + g t

v = (12.5 m/s) - (9.8 m/s²) (5.40 s)

v = (12.5 m/s) - (52.92 m/s)

v = - 40.4 m/s

(c) What total distance did it travel?

using equation of motion (iii), we have

v_f² = v₀² + 2 g d

(0)² = (12.5 m/s)² - 2 (9.8 m/s²) d                                                 

d = [(156.25 m²/s²) / (19.6 m/s²)]

d = 7.97 m

Then, we get

D_total = 2 d + y

D_total = [(15.94 m) + (75 m)]

D_total = 90.94 m

D_total$\approx$ 91 m