Question

Sheba, the American Pit Bull Terrier, runs and jumps off the dock with an initial speed of 9.12 m/s at an angle of 29° with respect to the surface of the water. (Assume that the +x axis is in the direction of the run and the +y axis is up.)(a) If Sheba begins at a height of 0.89 m above the surface of the water, through what horizontal distance does she travel before hitting the surface of the water?(b) Write an expression for the velocity of Sheba, in component form, the instant before she hits the water. (Express your answer in vector form.)(c) Determine the peak height above the water reached by Sheba during her jump.

Answer 1

a)The horizontal distance traveled was 8.53 m

b) The final velocity was v = (7.98 m/s, -6.06 m/s)

c) The peak height of Sheba above the water was 1.89 m

Step-by-step explanation:

The equations that describe the position and velocity in a parabolic motion are the following:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α +1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = vector position at time t

x0 = initial horizontal position

v0 = initial velocity

α = jumping angle

y0 = initial vertical position

t = time

g = acceleration due to gravity

v = vector position at time t

a) We have to find the x-component of the vector "r final" ( see figure). We know that the y-component of the vector "r final" is -0.89 m. We can use this data to find the time of flight and with that time we can calculate "rx final".

Then:

y = y0 + v0 · t · sin α +1/2 · g · t²

Since the origin of the frame of reference is located at the jump point, y0 = 0

y = v0 · t · sin α +1/2 · g · t²

-0.89 m = 9.12 m/s · t · sin 29° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 9.12 m/s · t · sin 29° + 0.89 m

Solving the quadratic equation:

t = 1.07 s (the negative value of t is discarded).

Now, we have the final time and can calculate rx final that is the horizontal displacement:

x = x0 + v0 · t · cos α (x0 = 0, same as y0)

x = 9.12 m/s · cos 29° · 1.07 s = 8.53 m

b) Using the equation for velocity and the final time calcualted above, we can obtain the velocity vector the instant Sheba hits the water:

v = (v0 · cos α, v0 · sin α + g · t)

vx = v0 · cos α

vx = 9.12 m/s · cos 29° = 7.98 m/s

vy = v0 · sin α + g · t

vy = 9.12 m/s · sin 29° - 9.8 m/s² · 1.07 s

vy = -6.06 m/s

The final velocity will be:

v = (7.98 m/s, -6.06 m/s) or v = 7.98 m/s i + (- 6.06 m/s j)

c) At the peak height, the y-component of the velocity is 0. We can use that to obtain the time and with that time we can calculate the vector position "r" (see figure).

vy = v0 · sin α + g · t

0 = v0 · sin α + g · t

-v0 · sin α/g = t

-9.12 · sin 29°/ -9.8 m/s² = t

t = 0.45 s

The y-component of the vector r, ry in the figure, will be:

y = y0 + v0 · t · sin α +1/2 · g · t² (y0 = 0)

y = 9.12 m/s · 0.45 s · sin 29° - 1/2 · 9.8 m/s² · (0.45 s)²

y = 0.997 m

The peak height of Sheba above the water was 0.997 m + 0. 89 m = 1.89 m