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

Launch speed is equal to
`19.11` meter/second

Step-by-step 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 = √(-2* a* S) \\`

Substituting the given values we get -

`U_0 * Sin 63 = √(- 2* -9.8* 14.8) \\U_0 = \frac{sqrt{- 2* -9.8* 14.8}}{Sin 63} \\U _0 = (17.031)/(0.891) \\U_0 = 19.11`

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