A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 15 m/s at an angl

Question

A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 15 m/s at an angle θ = 31 ° below horizontal. The stone moves without air resistance; use a Cartesian coordinate system with the origin at the stone’s initial position.

Required:
With what speed, vf in meters per second, does the stone strike the ground?

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Hưng Khoa 1 year 2021-09-04T13:16:03+00:00 1 Answers 416 views 0

Answers ( )

    -1
    2021-09-04T13:17:30+00:00

    Answer:

    The stone strikes the ground in 14.72m/s

    Explanation:

    The initial vertical velocity of the stone is given by:

    Vyo= VSin theta

    Given: V = 15m/s , theta = 31°

    Vyo = 15 Sin 31

    Vyo = 7.73m/s

    Final vertical velocity, Vf can be determined by using kinematic equation

    Vf^2 – U^2 = 2as

    Vf^2 -( 7.73)^2 = 2(9.8×(-8))

    Vf^2 = 59.75 + 156.8

    Vf^2 = 216.6

    Vf = sqrt(216.46)

    Vf = 14.72m/s

    Therefore,The stone strikes the ground in 14.72m/s

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