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A projectile is launched from ground level with an initial velocity of v_(0) feet per second. Neglecting air resistance, its height in feel i seconds after laut given by s=-16t^2+v_(0)t Find the time(s) that the projectile will (a) reach a height of 64 ft and (b) return to the ground when v_(0)=80 feet per second. (a) Find the time(s) that the projectile will reach a height of 64 ft when v_(0)=80 feel per second. Select the correct choice below and, if necessary, fill in answer box to complete your choice. OA. square seconds (Use a comma to separate answers as needed.) B. The projectile does not reach 64 feet (b) The projectile returns to the ground after square second(s)

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A projectile is launched from ground level with an initial velocity of v_(0) feet per second. Neglecting air resistance, its height in feel i seconds after laut given by s=-16t^2+v_(0)t Find the time(s) that the projectile will (a) reach a height of 64 ft and (b) return to the ground when v_(0)=80 feet per second. (a) Find the time(s) that the projectile will reach a height of 64 ft when v_(0)=80 feel per second. Select the correct choice below and, if necessary, fill in answer box to complete your choice. OA. square seconds (Use a comma to separate answers as needed.) B. The projectile does not reach 64 feet (b) The projectile returns to the ground after square second(s)

A projectile is launched from ground level with an initial velocity of v_(0) feet per second. Neglecting air resistance, its height in feel i seconds after laut given by s=-16t^2+v_(0)t Find the time(s) that the projectile will (a) reach a height of 64 ft and (b) return to the ground when v_(0)=80 feet per second. (a) Find the time(s) that the projectile will reach a height of 64 ft when v_(0)=80 feel per second. Select the correct choice below and, if necessary, fill in answer box to complete your choice. OA. square seconds (Use a comma to separate answers as needed.) B. The projectile does not reach 64 feet (b) The projectile returns to the ground after square second(s)

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ImaniProfessional · Tutor for 6 years

Answer

A. 1, 4 seconds<br />B. 5 seconds

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## Step 1: Formulate the equation for height <br />### Given \( s = -16 t^2 + v_0 t \).<br />### Plug in \( v_0 = 80 \): \( s = -16 t^2 + 80 t \).<br /><br />## Step 2: Solve for time when \( s = 64 \)<br />### Set \( -16 t^2 + 80 t = 64 \). <br />### Rearrange: \( -16 t^2 + 80 t - 64 = 0 \).<br /><br />## Step 3: Solve quadratic equation<br />### Use quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -16 \), \( b = 80 \), and \( c = -64 \).<br />### Calculate discriminant: \( b^2 - 4ac = 80^2 - 4(-16)(-64) = 6400 - 4096 = 2304 \).<br /><br />## Step 4: Find times <br />### Calculate \( t = \frac{-80 \pm \sqrt{2304}}{2(-16)} = \frac{-80 \pm 48}{-32} \).<br />### Solve for \( t = \frac{-80 + 48}{-32} \) and \( t = \frac{-80 - 48}{-32} \): \( t_1 = 1 \text{ sec} \), \( t_2 = 4 \text{ sec} \).<br /><br />## Step 5: Find time to return to ground <br />### Set \( -16 t^2 + 80 t = 0 \).<br />### Factor: \( t(-16 t + 80) = 0 \).<br />### Solve for \( t = 0 \) or \( t = 5 \text{ sec} \) (since \( t = 0 \) is initial launch time).<br /><br />#
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