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QuestionA dart is thrown at 11.6 m/s and hits the board in 0.131 s. Find the distance to the board and how far below the bullseye it hit.

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the dart is 11.6 m/s11.6 \mathrm{~m/s} . The dart hits the board after0.131 s3. The dart is thrown directly at the bullseye, so the initial vertical displacement is04. The acceleration due to gravity is 9.8 \mathrm{~m/s^} (downwards)
5. The dart follows a parabolic trajectory due to gravity

STEP 2

First, we need to find the horizontal distance to the board. Since there are no horizontal forces acting on the dart after it is thrown, it will maintain a constant horizontal velocity. We can find the horizontal distance by multiplying the horizontal velocity by the time it takes to hit the board.
Distance=VelocitytimesTimeDistance = Velocity \\times Time

STEP 3

Now, plug in the given values for the velocity and time to calculate the distance.
Distance=11.6 m/stimes0.131 sDistance =11.6 \mathrm{~m/s} \\times0.131 \mathrm{~s}

STEP 4

Calculate the distance to the board.
Distance=11.6 m/stimes0.131 s=1.52 mDistance =11.6 \mathrm{~m/s} \\times0.131 \mathrm{~s} =1.52 \mathrm{~m}

STEP 5

Next, we need to find how far below the bullseye the dart hit. We can do this by using the equation of motion for vertical displacement under constant acceleration (gravity).
Vertical displacement=Initial velocitytimesTime−0.5timesGravitytimesTime2Vertical\, displacement = Initial\, velocity \\times Time -0.5 \\times Gravity \\times Time^2

STEP 6

Since the dart is thrown directly at the bullseye, the initial vertical velocity is0. Plug in the values for time and gravity to calculate the vertical displacement.
Vertical displacement=0times0.131 s−0.5times9.8 m/s2times(0.131 s)2Vertical\, displacement =0 \\times0.131 \mathrm{~s} -0.5 \\times9.8 \mathrm{~m/s^2} \\times (0.131 \mathrm{~s})^2

STEP 7

Calculate the vertical displacement.
Vertical displacement=0−0.5times9. m/s2times(0.131 s)2=−0.084 mVertical\, displacement =0 -0.5 \\times9. \mathrm{~m/s^2} \\times (0.131 \mathrm{~s})^2 = -0.084 \mathrm{~m}The negative sign indicates that the dart hit below the bullseye.
a. The board is1.52 m away. b. The dart hit0.084 m below the bullseye.

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