Question

A cork shoots out of a champagne bottle at an angle of 40.0 above the horizontal. If the cork travels a horizontal distance of 1.50 m in 1.25 s, what was its initial speed?

Answer 1

Answer:

The initial speed of the cork was 1.57 m/s.

Explanation:

Hi there!

The equation of the horizontal position of the cork in function of time is the following:

x = x0 + v0 · t · cos θ

Where:

x = horizontal position at time t.

x0 = initial horizontal position.

v0 = initial speed of the cork.

t = time.

θ = launching angle.

If we place the origin of the frame of reference at the launching point, then x0 = 0.

We know that at t = 1.25 s, x = 1.50 m. We also know the launching angle so we can solve the equation of horizontal position for the initial speed, v0:

x = v0 · t · cos θ

x / t · cos θ = v0

v0 = 1.50 m / (1.25 s · cos (40.0°)

v0 = 1.57 m/s

The initial speed of the cork was 1.57 m/s.

Answer 2

Answer:

1.57 m/s.

Explanation:

Horizontal velocity = distance/time

= 1.5/1.25

= 1.2 m/s

Vo * CosΘ = Vx

Cos(40) = 1.2/x

= 1.2/cos(40)

= 1.57 m/s.