Answer: 7.78m/s
Explanation: As the the skier slide down the height, we assume the motion of a body, slidind down an incline plane.
Force down the plane= [email protected]
Frictional force= umg
u= coefficient of friction
Net force on skier = [email protected] umg
ma = [email protected]
a = g([email protected] - u) = 9.8 (sin 25- 0.2)
a = 9.8 × (0.4226-0.2) = 9.8 × 0.2226
a = 2.18m/s²
Using the formula V² = U² + 2aH
Where H = (10.4+ 3.5)=total height of descent before landing, U= 0.
V = √ 2 × 2.18× 13.9 = √60.604
V = 7.78m/s
Assuming the object was originally at rest, it must have been traveling for
14.0/3.45 = 4.06 seconds
Vector trigonometry can be used for this problem. Since the horizontal component is 12 meters per second, this is technically the hypotenuse (actual initial velocity) multiplied to cosine of 40 degrees. Therefore, to find the hypotenuse, we must divide 12 by cosine 40degrees. cos(40)= 0.766, and 12/0.766 = approximately 15.664, therefore our answer is (3) 15.7 m/s
Answer:
Explanation:
the variations in riser height or tread depth should not be grater than that is equal to 9.5 mm but the maximum riser height should be the 
but variation in riser height should not exceed to . The minimum riser height should be 7 inches which is equal to the 178 mm
