Physics help required ?

A projectile starts from rest and moves 4.8 m

down a frictionless ramp inclined at 14◦ with

the horizontal.

The acceleration of gravity is 9.8 m/s ^2

With what speed will it leave the ramp?

Answer in units of m/s.

Part 2

What will be the range of the projectile if

the bottom of the ramp is 2.2 m above the

ground?

Answer in units of m.

1 Answer

Relevance
  • Whome
    Lv 7
    1 month ago
    Favorite Answer

    v = √(2(-9.8)(4.8sin-14) = 4.77 m/s

    -2.2 = 0 + (4.77sin-14)t + ½(-9.8)t²

    0 = 4.9t² + 1.15t - 2.2

    quadratic formula or other means gives a positive answer of 

    t = 0.56 s

    r = (4.77cos-14)(0.56) = 2.6 m

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