(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

(a) The height of the cliff is 9.92 m

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

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;

Time to reach the maximum height;

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;

the average speed is 24.57 km/hr

double the distance = divide strength by 4