Answer:
[tex]v_0=20.7\frac{m}{s}[/tex]
Explanation:
We are given the equation for the height:
[tex]h(t)=10+v_0t-4.9t^2[/tex]If we differentiate this equation, we get the equation of the velocity:
[tex]h^{\prime}(t)=v(t)=v_0-9.8t[/tex]The problem tells us that the object hits the ground at a velocity of -25m/s. We can write this:
[tex]-25=v_0-9.8t_0[/tex]Where t0 is the time when the object hits the ground, and the velocity is -25m/s
Now, we can solve for v0:
[tex]v_0=9.8t_0-25[/tex]And if we use this in the height h(t) equation, we can find the value of t0. At t0, the height is 0. Thus:
[tex]0=10+(9.8t_0-25)t_0-4.9t_0^2[/tex]And solve:
[tex]\begin{gathered} 0=10+9.8t_o^2-4.9t_0^2-25t_0 \\ 0=10-25t_0+4.9t_0^2 \end{gathered}[/tex]Next, we can use the quadratic formula to solve this:
[tex]\begin{gathered} t_{\pm}=\frac{-(-25)\pm\sqrt{(-25)^2-4\cdot4.9\cdot10}}{2\cdot4.9}=\frac{25\pm\sqrt{625-196}}{9.8}=\frac{25\pm\sqrt{429}}{9.8}=\frac{25\pm20.71231}{9.8} \\ . \\ t_+=\frac{25+20.71231}{9.8}=4.664521seconds \\ . \\ t_-=\frac{25-20.71231}{9.8}=0.43751seconds \end{gathered}[/tex]Let's see which of the two solutions is the time we are looking for.
Let's go back to our solution for v0:
[tex]v_0=9.8t_0-25[/tex]If we use t0 = 0.43751 s:
[tex]v_0=9.8\cdot0.46751-25=-20.7123m/s[/tex]This means that the initial velocity is negative, thus the object goes downwards. But, the problem tells us that initially the object is thrown upwards.
The correct value of t0 = 4.66 seconds
Now we can find v0:
[tex]v_0=9.8\cdot4.664521-25=20.712[/tex]Thus, v0 = 20.7 m/s
