# (need answer in 2 days please).You throw a javelin at a speed of 23 m/s at an angle of 40° above the horizontal.?

You throw a javelin at a speed of 23 m/s at an angle of 40° above the horizontal.

a. How far does the javelin travel vertically? (2 points)

b. You threw the javelin from an original height of 2 m, but you accidentally threw it off a cliff that drops an additional 20 m. How long does it take the javelin to reach the ground? (4 points)

c. How far does the javelin travel horizontally from where you threw it? (4 points)

Relevance

Take g = 9.81 m/s²

Take all upward quantities to be positive, and downward to be negative.

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a.

Consider the vertical motion (uniform acceleration motion):

Initial velocity, u(y) = 23 sin40° m/s

Final velocity, v(y) = 0 m/s

Acceleration, a(y) = -9.81 m/s²

v(y)² = u(y)² + 2 a(y) s(y)

0 = (23 sin40⁰)² + 2 (-9.81) s(y)

Vertical distance traveled, s(y) = (23 sin40⁰)² / (2 * 9.81) m = 11 m

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b.

Refer to the diagram below for parts b and c.

Consider the vertical motion (uniform acceleration motion):

Initial velocity, u(y) = 23 sin40° m/s

Displacement, s(y) = -(2 + 20) = -22 m/s

Acceleration, a(y) = -9.81 m/s²

s(y) = u(y) t + (1/2) a(y) t²

-22 = (23 sin40°) t + (1/2) (-9.81) t²

4.905t² - 14.78t - 22 = 0

t = [14.78 ± √(14.78² + 4*4.095*22)] / (2*4.095)

t = 4.1 s or t = -1.1 s (rejected)

Time taken = 4.1 s

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c.

Consider the horizontal motion (uniform velocity motion):

Velocity, v(x) = 23 cos40° m/s

Time taken, t = 4.1 s

s(x) = v(x) t = (23 cos40°) * 4.11 = 72 m

Horizontal distance traveled = 72 m • Anonymous
8 months ago

a)

Voy = Vo*sin 40° = 23*0.643 = 14.78 m/sec

h = Voy^2/2g = 14.78^2/19.612 = 11.14 m

b)

-(20+2) = Voy*t-g/2*t^2

-22-14.78t+4.903t^2 = 0

t = (14.78+√

14.78^2+19.612*22)/9.806 = 4.107 sec

c)

d = Vo*cos 40*t = 23*0.766*4.107 ≅ 72.36 m