Question

In the aerials competition in skiing, the competitors 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. The end of a launch ramp is directed 63° above the horizontal. With this launch angle, a skier attains a height of 14.8 m above the end of the ramp. What is the skier’s launch speed?

Answer 1

Answer:

Launch speed is equal to $19.11$ meter/second

Explanation:

Using the newton's equation, we know that

$V^2 = U^ 2 + 2 *a*S$

Where V is the final speed , U is the initial speed, a is the acceleration and S is the distance

At the highest point of the trajectory the object;s speed reduces to zero. hence

$V = 0$

While initial velocity is equal to $U sin 63$°

$U^2 = -2* a* S\\U = \sqrt{-2* a* S}$

Substituting the given values we get -

$U_0 * Sin 63 = \sqrt{- 2* -9.8* 14.8} \\U_0 = \frac{sqrt{- 2* -9.8* 14.8}}{Sin 63} \\U _0 = \frac{17.031}{0.891} \\U_0 = 19.11$

Thus, launch speed is equal to $19.11$ meter/second