physics...projectile velocity and kinematics in two dimensions question?

Question

The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. In the women's competition, the end of a typical launch ramp is directed 58° above the horizontal. With this launch angle, a skier attains a height of 14 m above the end of the ramp. What is the skier's launch speed?

Answer 1

You need to start by determining what vertical velocity results in a maximum height of 14 m under the influence of gravity. Once you have this initial vertical velocity, v_0y, you can use trigonometry to determine the initial velocity on the 58° ramp.

In projectile motion, vertical and horizontal velocity can be treated separately. Looking at just the vertical velocity, maximum height is achieved at the instant that vertical velocity becomes zero. After that, the object begins moving downward. With an intial vertical velocity of v_0y, it takes a time equal to (v_0y)/g for gravity to slow the vertical motion to zero. The equation for distance traveled, given initial velocity v_0 and acceleration a, is x = (v_0)*t + 0.5*a*t^2. In this case, that becomes y = (v_0y)*(v_0y)/g - 0.5*g*[(v_0y)/g]^2, where we take g = 9.8 m/s^2, a positive value. y was given as 14 m, so we have 14 = (v_0y)^2/9.8 - 0.5*(v_0y)^2/9.8 = 0.5*(v_0y)^2/9.8 ==> v_0y = sqrt(9.8*14*2), and I leave the final calculation to you.

Now that you have v_0y, the rest is just trigonometry. The actual launch velocity, v_0, is the hypotenuse of a right triangle where v_0y is the leg opposite an angle of 58°. That means that v_0*sin(58°) = v_0y, and v_0y is now known, so you can solve algebraically for v_0, the launch speed.

Answer 2

horizontal area of velocity = 19.5 cos 32.8 = sixteen.39 m/s This remains consistent as gravity acts in the vertical plane (assuming 0 air resistance too) Time to attain goalie = 28.2/sixteen.39 = a million.seventy two s Vertical area of velocity = 19.5 sin32.8 = 10.fifty six m/s it truly is concern to gravity, so after a million.seventy two sec the vertical velocity is: v=u+at = 10.fifty 9 - 9.8*a million.seventy two = -6.27 m/s (minus exhibits its moving downwards). Now use pythag to discover blended velocity of horizontal and vertical compnents. velocity^2=-6.27^2 + sixteen.39^2 velocity = 17.fifty 5 m/s