Question

An object is thrown vertically upward such that it has a speed of 75 m/s when it reaches two thirds of its maximum height above the launch point. Determine the maximum height h. The acceleration of gravity is 9.8 m/s² .
(a)96m
(b)32m
(c)48 m
(d)75m
(e)64m

Answer 1

The maximum height is equal to 860.04m

Why?

We can find the maximum height of the object in two steps.

First step: Finding the initial speed.

$v_{f}^{2}=v_{o}^{2}-2*g*d\\\\0=v_{o}^{2}-2*(9.81\frac{m}{s^{2}})*hmax\\\\hmax=\frac{v_{o}^{2}}{2*9.81\frac{m}{s^{2} }}$

Then,

$v_{f}^{2}=v_{o}^{2}-2*g*d\\\\(75\frac{m}{s})^{2} =v_{o}^{2}-2*(9.81\frac{m}{s^{2}})*\frac{2}{3}\frac{v_{o}^{2}}{2*9.81\frac{m}{s^{2} }}\\\\5625\frac{m^{2} }{s^{2}}=v_{o}^{2}-\frac{2}{3}v_{o}^{2}\\\\3*5625\frac{m^{2} }{s^{2}}=v_{o}^{2}\\\\v_{o}=\sqrt{3*5625\frac{m^{2} }{s^{2}}}=129.90\frac{m}{s}$

Second step: Find the maximum height.

Now, again using the second equation, we need to subsitutite the obtained value for the initial speed into it, so:

$hmax=\frac{(129.90\frac{m}{s})^{2}}{2*9.81\frac{m}{s^{2} }}=860.04m$

Hence, we have that the maximum height is equal to 860.04 meters.

Have a nice day!