Question

Chapter 03, Problem 75 As preparation for this problem, review Conceptual Example 9. From the top of a cliff overlooking a lake, a person throws two stones, as shown in the drawing. The cliff is 23.6 m high. The two stones described have identical initial speeds of vo-13.5 m/s and are thrown at an angle θ-26.5。, one below the horizontal and one above the horizontal, what is the distance between the points where the stones strike the water? Neglect air resistance Number Units the tolerance is +/-2%

Answer 1

here,

the height of cliff , h = 23.6 m

v0 = 13.5 m/s

theta = 26.5 degree

let the time taken to hit the ground be t1

for first stone

- h = v0 * t1 * sin(26.5) - 0.5 * g * t1^2

- 23.6 = 13.5 * t1 * sin(26.5) - 0.5 * 9.81 * t1^2

solving for t1

t1 = 2.89 s

let the time taken to hit the ground be t2

for second stone

- h = - v0 * t2 * sin(26.5) - 0.5 * g * t2^2

- 23.6 = - 13.5 * t2 * sin(26.5) - 0.5 * 9.81 * t2^2

solving for t2

t2 = 1.66 s

the difference between the horizontal distance , x = v0 * cos(theta) * ( t1 - t2)

x = 13.5 * cos(26.5) * ( 2.89 - 1.66)

x = 14.86 m