Answer to Question #250719 in Mechanics | Relativity for Gold

Question

With what velocity in m/s must an object be thrown to reach a maximum height of 90.6m? Assume negligible air resistance, indicate proper sign convention

Answer 1

We have the system where the velocity has to be directed to the y+ section, and the acceleration a = - g = - 9.80 m/s² because it is against the displacement.

$\text {At the highest altitude, velocity for the y-axis is zero:} \\V_y=V_{oy}-gt=0 \implies t= \cfrac{V_{0y}}{g} \\ \text{On the other hand, the maximum height will be found} \\ \text{ when we substitute t on the equation for H:} \\ H=V_{oy}\cdot t-\frac{1}{2}gt^2= t(V_{oy}-\frac{1}{2}gt) \\ H= \Big(\frac{V_0}{g}\Big) \Big(V_0-\frac{g}{2} \big(\frac{V_0}{g}\big) \Big) \\ \therefore H=\cfrac{V^2_0}{2g} \implies V_0=\sqrt{2gH}$

We substitute the height that is supposed to be reached (H = 90.6 m) and then we find the initial velocity as $V_0=\sqrt{2(9.80\frac{m}{s})(90.6\,m)}=42.14\frac{m}{s}$.

In conclusion, the object has to be thrown with a velocity of 42.14 m/s to be able to reach H = 90.6 m.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.