Question

(a) Calculate the height (in m) of a cliff if it takes 2.48 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.19 m/s. m (b) How long (in s) would it take to reach the ground if it is thrown straight down with the same speed

Answer 1

(a) The height of the cliff is 9.92 m

(b) Time taken to reach the ground is 0.81 s

Step-by-step explanation:

Given;

time of motion of the rock, t = 2.48 s

initial velocity of the rock, u = 8.19 m/s

(a) This is a two-way problem;

maximum upward height reached by the rock (h) + Total distance traveled downwards from the top (H).

Height of cliff = H - h

Maximum height reached by the rock;

`v^2 = u^2 - 2gh\\\\0 = (8.19)^2 - (2* 9.8)h\\\\0 = 67.08 - 19.6 h\\\\19.6h = 67.08\\\\h = (67.08)/(19.6) \\\\h = 3.422 \ m`

Time to reach the maximum height;

`h = ut - (1)/(2) gt^2\\\\3.422 = 8.19t - (0.5 * 9.8)t^2\\\\3.422 = 8.19t - 4.9t^2\\\\4.9t^2 -8.19t + 3.422 = 0\\\\This \ forms \ a quadratic \ equation ; \\\\a = 4.9, b = -8.19, c = 3.422\\\\t = (-b \ \ +/- \ \ √(b^2 -4ac) )/(2a) \\\\t = (-(-8.19) \ \ +/- \ \ √((-8.19)^2 -4(4.9* 3.422)) )/(2(4.9)) \\\\t = (8.19 \ \ + /- \ \ √(0.0049) )/(9.8) \\\\t = (8.19 - 0.07)/(9.8) \ \ or \ \ t = (8.19 + 0.07)/(9.8) \\\\t = 0.83 \ s \ \ or \ \ t = 0.84 \ s\\\\`

The time taken for rock to falll to the gound;

t₂ = 2.48 - 0.83 s

t₂ = 1.65 s

The distance traveled downwards is calculated as

H = vt + ¹/₂gt²

where;

v is the final velocity at maximum height = initial velocity when the rock starts moving downwards = 0

H = ¹/₂gt²

H = 0.5 x 9.8 x (1.65)²

H = 13.34 m

The height of the cliff = 13.34 m - 3.422m = 9.92 m

(b) time taken to reach the ground if thrown with the same speed;

h = ut + ¹/₂gt²

9.92 = 8.19t + (0.5 x 9.8)t²

9.92 = 8.19t + 4.9t²

4.9t² + 8.19t - 9.92 = 0

a = 4.9, b = 8.19, c = -9.92

use the quadratic formula;

`t = (-b \ \ +/- \ \ √(b^2 -4ac) )/(2a) \\\\t = (-8.19 \ \ +/- \ \ √((8.19)^2 -4(4.9* -9.92)) )/(2(4.9)) \\\\t = 0.81 \ s`