A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing. (a) Neglecting air resistance, how long does it take the skier to reach the ground?

(b) How far horizontally does the skier travel in the air before landing?

m

(a).The time taken by the skier to reach the ground is 0.145 second.

(b).The skier travel in the air before landing is 19.29 meter.

a. Given that A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.

Using equation of motion.

Where S is vertical distance , u is initial velocity and g is gravitational acceleration.

Substitute S = 3.45 m, u = 23m/s  and g = 9.8 in above equation.

Since, time can not be negative.

So that,  

b. The horizontal distance travel before landing is known as Range.

           Horizontal distance  ,

Substitute v = 23m/s , S = 3.45 and g = 9.8 meter per second square.

Thus, The skier travel in the air before landing is 19.29 meter.

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Explanation:

Given

Velocity v = 23.0m/s

Distance S = 3.45m

Required

Time it will take the skier to reach the ground;

Using the equation of motion;

S = ut + 1/2gt²

3.45 = 23t + 1/2(9.8)t²

3.45 = 23t + 4.9t²

4.9t²+23t-3.45 = 0

Factorize;

t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)

t = -23 ±√529+67.62/9.8

t = -23±√596.62/9.8

t = -23±24.43/9.8

t = 1.43/9.8

t = 0.146 secs

Hence take the skier 0.146 secs to reach the ground.

b) Horizontal distance covered is the range;

Range = U√2H/g

Range = 23√2(3.45)/9.8

Range = 23√6.9/9.8

Range = 23√0.7041

Range = 23(0.8391)

Range = 19.29m

Hence the horizontal distance travelled in air is 19.29m

double the distance = divide strength by 4