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A projectile is launched straight up from the ground with an initial velocity of 120ft/s If acceleration due to gravity is - 16ft/s^2 after about how many seconds will the object reach a height of 200 ft? h(t)=at^2+vt+h_(0) 1:0 s 1.7 s 2.5 s 8.0 s

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A projectile is launched straight up from the ground with an initial velocity of 120ft/s If acceleration due to gravity is - 16ft/s^2 after about how many seconds will the object reach a height of 200 ft? h(t)=at^2+vt+h_(0) 1:0 s 1.7 s 2.5 s 8.0 s

A projectile is launched straight up from the ground with an initial velocity of 120ft/s If acceleration due to gravity is - 16ft/s^2 after about how many seconds will the object reach a height of 200 ft? h(t)=at^2+vt+h_(0) 1:0 s 1.7 s 2.5 s 8.0 s

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EllaProfessional · Tutor for 6 years

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### \( 2.5 \text{ s} \)

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## Step1: Identify the known and unknown variables<br />### The given initial velocity is \( v_0 = 120 \, \text{ft/s} \), the height to reach is \( h = 200 \, \text{ft} \), and the acceleration due to gravity is \( a = -16 \, \text{ft/s}^2 \). We need to determine the time \( t \).<br /><br />## Step2: Write the height equation<br />### The height equation for projectile motion is \( h(t) = a t^2 + v_0 t + h_0 \). Here \( h_0 = 0 \) since the object is launched from the ground level.<br /><br />## Step3: Substitute the known values into the height equation<br />### Substitute \( h = 200 \, \text{ft} \), \( a = -16 \, \text{ft/s}^2 \), \( v_0 = 120 \, \text{ft/s} \), and \( h_0 = 0 \) into the equation:<br />\[<br />200 = -16 t^2 + 120 t<br />\]<br /><br />## Step4: Solve the quadratic equation<br />### Rearrange the equation to standard quadratic form:<br />\[<br />16 t^2 - 120 t + 200 = 0<br />\]<br />Divide the entire equation by 4 for simplicity:<br />\[<br />4 t^2 - 30 t + 50 = 0<br />\]<br />Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4 \), \( b = -30 \), and \( c = 50 \):<br /><br />Calculate the discriminant: <br />\[<br />\Delta = b^2 - 4ac = (-30)^2 - 4(4)(50) = 900 - 800 = 100<br />\]<br /><br />Find the roots:<br />\[<br />t = \frac{30 \pm \sqrt{100}}{8} = \frac{30 \pm 10}{8}<br />\]<br />This gives two solutions:<br />\[<br />t = \frac{40}{8} = 5 \, \text{s}, \quad t = \frac{20}{8} = 2.5 \, \text{s}<br />\]<br /><br />## Step5: Select the feasible time<br />### Since 5 seconds seems impractically long for the object to ascend and still be rising, select the smaller, realistic time value.<br /><br />#
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