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11. [0/1 Points] DETAILS PREVIOUS ANSWERS SERCP11 3.2.OP.013. 3/10 Submissions Used MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A rock is tossed from the top of a building at an angle of 10° above the horizontal with an initial speed of 17 m/s. The rock lands on the ground 2.9 s after it is tossed. What is the height of the building (in m)? 32.6 X What constant-acceleration formula relates the vertical displacement to initial velocity, acceleration, and time? How is the initial vertical component of the velocity related to the initial speed and angle? What is the direction of the initial vertical component of velocity, and what is the direction of the acceleration? m Solution or Explanation We choose our origin at the initial position of the projectile. After 2.9 s, it is at ground level, so the vertical displacement is \(\Delta y = -H\). To find H, we use the following equation. \(\Delta y = v_{0y}t + \frac{1}{2}a_yt^2\) We then plug in the information we have to find H. \(-H = (17 \text{ m/s})(\sin(10°))(2.9 \text{ s}) + \frac{1}{2}(-9.80 \text{ m/s}^2)(2.9 \text{ s})^2\) or \(H = 32.6 \text{ m}\) Need Help? Read It Show My Work (Optional) Submit Answer

          11. [0/1 Points] DETAILS PREVIOUS ANSWERS SERCP11 3.2.OP.013. 3/10 Submissions Used MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A rock is tossed from the top of a building at an angle of 10° above the horizontal with an initial speed of 17 m/s. The rock lands on the ground 2.9 s after it is tossed. What is the height of the building (in m)? 32.6 X What constant-acceleration formula relates the vertical displacement to initial velocity, acceleration, and time? How is the initial vertical component of the velocity related to the initial speed and angle? What is the direction of the initial vertical component of velocity, and what is the direction of the acceleration? m Solution or Explanation We choose our origin at the initial position of the projectile. After 2.9 s, it is at ground level, so the vertical displacement is \(\Delta y = -H\). To find H, we use the following equation. \(\Delta y = v_{0y}t + \frac{1}{2}a_yt^2\) We then plug in the information we have to find H. \(-H = (17 \text{ m/s})(\sin(10°))(2.9 \text{ s}) + \frac{1}{2}(-9.80 \text{ m/s}^2)(2.9 \text{ s})^2\) or \(H = 32.6 \text{ m}\) Need Help? Read It Show My Work (Optional) Submit Answer

11. [0/1 Points] DETAILS PREVIOUS ANSWERS SERCP11 3.2.OP.013. 3/10 Submissions Used MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A rock is tossed from the top of a building at an angle of 10° above the horizontal with an initial speed of 17 m/s. The rock lands on the ground 2.9 s after it is tossed. What is the height of the building (in m)? 32.6 X What constant-acceleration formula relates the vertical displacement to initial velocity, acceleration, and time? How is the initial vertical component of the velocity related to the initial speed and angle? What is the direction of the initial vertical component of velocity, and what is the direction of the acceleration? m Solution or Explanation We choose our origin at the initial position of the projectile. After 2.9 s, it is at ground level, so the vertical displacement is Δ y = -H. To find H, we use the following equation. Δ y = v0yt + (1)/(2)ayt^2 We then plug in the information we have to find H. -H = (17 m/s)(sin(10°))(2.9 s) + (1)/(2)(-9.80 m/s^2)(2.9 s)^2 or H = 32.6 m Need Help? Read It Show My Work (Optional) Submit Answer

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