Respuesta :
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
(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.
[tex]S=ut+\frac{1}{2}gt^{2}[/tex]
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.
[tex]3.45=23t+\frac{1}{2} (9.8)t^{2}\\\\4.9x^{2} +23t-3.45=0\\\\t=0.145,-4.83[/tex]
Since, time can not be negative.
So that, [tex]t=0.145s[/tex]
b. The horizontal distance travel before landing is known as Range.
Horizontal distance ,
[tex]=v*\sqrt{\frac{2S}{g} }[/tex]
Substitute v = 23m/s , S = 3.45 and g = 9.8 meter per second square.
[tex]Distance=23*\sqrt{\frac{2*3.45}{9.8} } \\\\Distance=23*\sqrt{0.7041} \\\\Distance=23*0.8391=19.29m[/tex]
Thus, The skier travel in the air before landing is 19.29 meter.
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