QuestionMarch 21, 2026

That previous shot was a bust and the player tries for a second shot. This time they throw the ball at 8m/s with a 45 degree angle. The height of the throw is 2.4 m above the floor and the hoop is 3.1 m tall. How long will the ball be in the air before reaching the height of the hoop? (hint:we've done this in examples and there are two ways of doing it) 8.11 sec 0.63 sec 0.12 sec 1.01 sec 1.35 sec 5.31 sec

That previous shot was a bust and the player tries for a second shot. This time they throw the ball at 8m/s with a 45 degree angle. The height of the throw is 2.4 m above the floor and the hoop is 3.1 m tall. How long will the ball be in the air before reaching the height of the hoop? (hint:we've done this in examples and there are two ways of doing it) 8.11 sec 0.63 sec 0.12 sec 1.01 sec 1.35 sec 5.31 sec
That previous shot was a bust and the player tries for a second shot. This time they throw the ball at 8m/s with a 45 degree angle. The height of the throw is 2.4 m above the floor and the hoop is 3.1 m tall. How long will the ball be in the air before reaching the height of the hoop? (hint:we've done this in examples and there are two ways of doing it) 8.11 sec 0.63 sec 0.12 sec 1.01 sec 1.35 sec 5.31 sec

Solution
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Answer

1.01 sec Explanation 1. Break velocity into components v_y = v_0 \sin 45^\circ = 8 \times \frac{\sqrt{2}}{2} \approx 5.657\ \text{m/s} 2. Use vertical motion equation Formula: **y = y_0 + v_y t - \frac{1}{2} g t^2** We set y = 3.1, y_0 = 2.4, v_y \approx 5.657, g = 9.8: 3.1 = 2.4 + 5.657 t - 4.9 t^2 3. Rearrange equation 0.7 = 5.657 t - 4.9 t^2 4.9 t^2 - 5.657 t + 0.7 = 0 4. Solve quadratic t = \frac{5.657 \pm \sqrt{(5.657)^2 - 4(4.9)(0.7)}}{2 \times 4.9} t = \frac{5.657 \pm \sqrt{32.000 - 13.72}}{9.8} t = \frac{5.657 \pm \sqrt{18.28}}{9.8} t = \frac{5.657 \pm 4.276}{9.8} 5. Select positive times t_1 \approx \frac{9.933}{9.8} \approx 1.01\ \text{s} t_2 \approx \frac{1.381}{9.8} \approx 0.141\ \text{s} Only upward time to hoop is relevant for first hit → t \approx 1.01\ \text{s}

Explanation

1. Break velocity into components <br /> $v_y = v_0 \sin 45^\circ = 8 \times \frac{\sqrt{2}}{2} \approx 5.657\ \text{m/s}$ <br /><br />2. Use vertical motion equation <br /> Formula: **$y = y_0 + v_y t - \frac{1}{2} g t^2$** <br />We set $y = 3.1$, $y_0 = 2.4$, $v_y \approx 5.657$, $g = 9.8$: <br />$3.1 = 2.4 + 5.657 t - 4.9 t^2$ <br /><br />3. Rearrange equation <br /> $0.7 = 5.657 t - 4.9 t^2$ <br /> $4.9 t^2 - 5.657 t + 0.7 = 0$ <br /><br />4. Solve quadratic <br /> $t = \frac{5.657 \pm \sqrt{(5.657)^2 - 4(4.9)(0.7)}}{2 \times 4.9}$ <br /> $t = \frac{5.657 \pm \sqrt{32.000 - 13.72}}{9.8}$ <br /> $t = \frac{5.657 \pm \sqrt{18.28}}{9.8}$ <br /> $t = \frac{5.657 \pm 4.276}{9.8}$ <br /><br />5. Select positive times <br /> $t_1 \approx \frac{9.933}{9.8} \approx 1.01\ \text{s}$ <br /> $t_2 \approx \frac{1.381}{9.8} \approx 0.141\ \text{s}$ <br />Only upward time to hoop is relevant for first hit → $t \approx 1.01\ \text{s$}
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