Answer :
Let's first find the velocity in the x-direction which is constant.
[tex]d = vt \\ v = \frac{d}{t} \\v = \frac{83.2m}{4.12s} \\ v= 20.2m/s[/tex]
Now, we need the velocity in the y-direction
[tex]d = v_{o}t + \frac{1}{2}a t^{2} \\ d - \frac{1}{2}at^{2} = v_{o}t\\v_{o} = \frac{0 - \frac{1}{2}(-9.81)(4.12)^{2}}{4.12}\\v_{o} = 20.2m/s[/tex]
Finally, we need to put the two velocities together using pythagorean theorem
[tex]v_{t}^{2} = v_{x}^{2} + v_{y}^{2}[/tex]
[tex]v_{t} = \sqrt{20.2^{2} + 20.2^{2}}[/tex]
v = 28.6m/s
[tex]d = vt \\ v = \frac{d}{t} \\v = \frac{83.2m}{4.12s} \\ v= 20.2m/s[/tex]
Now, we need the velocity in the y-direction
[tex]d = v_{o}t + \frac{1}{2}a t^{2} \\ d - \frac{1}{2}at^{2} = v_{o}t\\v_{o} = \frac{0 - \frac{1}{2}(-9.81)(4.12)^{2}}{4.12}\\v_{o} = 20.2m/s[/tex]
Finally, we need to put the two velocities together using pythagorean theorem
[tex]v_{t}^{2} = v_{x}^{2} + v_{y}^{2}[/tex]
[tex]v_{t} = \sqrt{20.2^{2} + 20.2^{2}}[/tex]
v = 28.6m/s
