Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an angle of 45

Question

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an angle of 45 degrees. The ball is 2.0 m off the ground when it leaves his hand, and hits the backboard 3.5 m off the ground. Neglecting air friction, about how long is the ball in flight?

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Khoii Minh 5 years 2021-09-04T12:12:11+00:00 1 Answers 49 views 0

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  1. minhkhang
    0
    2021-09-04T12:14:05+00:00
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    Answer:

    The time of flight of the ball is 1.06 seconds.

    Explanation:

    Given \Delta x=7\ m

    \theta=45 \°

    Also, \Delta y=(3.5-2)=1.5\ m

    a_x=0\ and\ a_y=-9.81\ m/s^2

    Let us say the velocity in the x-direction is v_x and in the y-direction is v_y. And acceleration in the x-direction is a_x and in the y-direction is a_y.

    Also, \Delta x\ and\ \Delta y is distance covered in x and y direction respectively. And t is the time taken by the ball to hit the backboard.

    We can write v_x=v_0cos(45)\ and\ v_y=v_0sin(45). Where v_0 is velocity of ball.

    Now,

    \Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt

    \Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}

    Also,

    \Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s.

    Plugging this value in

    t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}

    t=1.06\ seconds

    So, the time of flight of the ball is 1.06 seconds.

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