A ball is thrown vertically upward, how long would it take to reach 300m below it's starting point if the initial velocity is 32m/s. What is the final velocity the moment it reaches that point?
Answer (1)
It would take approximately 11.74 seconds to reach 300m below its starting point, and the final velocity at that moment would be approximately -83.1 m/s (downwards).Data givenInitial velocity ([tex]v_{i}[/tex]) = +32 m/s (positive because it's thrown upward)Displacement (Δ[tex]y[/tex]) = -300 m (negative because the final position is 300m below the starting point)Acceleration due to gravity (a) = -9.8 m/s² (negative because gravity pulls downward)Time (t) = ?Final velocity ( [tex]v_{f}[/tex]) = ?To solve for the velocity, we will use the kinematic equationΔ[tex]y = v_{i} t + \frac{1}{2} at^{2}[/tex][tex]-300 = (32)t + \frac{1}{2}(-9.8)t^{2}\\ -300 = 32t - 4.9t^{2}[/tex]Rearrange this into a standard quadratic equation ( [tex]at^{2} + bt + c = 0[/tex] ).[tex]4.9t^{2} - 32t - 300 = 0[/tex]Now, use the quadratic formula to solve for t.(Please note that the operator symbol after (-32) is ±)[tex]t = \frac{-(-32)+ \sqrt{(-32)^{2}-4(4.9)(-300) } }{2(4.9)} \\t = \frac{32+\sqrt{1024+5880} }{9.8} \\t = \frac{32+\sqrt{6904} }{9.8}\\ t = \frac{32+83.09}{9.8}[/tex]We will now get two possible values for t.[tex]t_{1} = \frac{32+83.09}{9.8} = \frac{115.09}{9.8}[/tex] ≈ 11.744 seconds[tex]t_{2} = \frac{32 - 83.09}{9.8} = \frac{-51.09}{9.8}[/tex] ≈ -5.21 secondsSince time cannot be negative, we choose the positive value.Time (t) ≈ 11.74 secondsNow that we have the value of time, we can use the kinematic equation below.[tex]v_{f} = v_{i} + at\\v_{f} = 32 + (-9.8)(11.74)\\v_{f} = - 83.05 m/s[/tex]Rounded off this is -83.1 meters per second.
