Question

a ball thrown upwards returns to the ground with the initial speed of u. the ball is at a height of 80 m at two times, the time interval being 6 s. find u by assuming g=10 m/s

Answer 1

Answer on Question#55168 – Physics – Mechanics | Kinematics | Dynamics

Question

A ball thrown upwards returns to the ground with the initial speed of $u$. The ball is at a height of $80 \, m$ at two times, the time interval being $6 \, s$. Find $u$ by assuming $g = 10 \frac{m}{s}$.

Solution

"Ball returns to the ground with the initial speed", hence there is no friction.

If there is no friction, we can use the conservation law: $mgh = \frac{mv^2}{2}$, where $m$ – mass of the ball, $h$ – highest position of the ball, $v \equiv u$ – initial speed of the ball.

Due to the symmetry of process (once again, because there is no friction) it takes the same time for the ball to move upward from $80 \, m$ to $h$, as to move downward from $h$ to $80 \, m$.

Thus, we can formulate next equation:

$$h - 80 = \frac{g * 3^2}{2}$$

It comes from general formula (free fall): $s = \frac{gt^2}{2} + v_0 t$, we consider covered distance ($s$) to be the distance between the highest point and $80 \, m$, respective time according to the statement above is $\frac{6}{2} = 3 \, s$, we consider start of movement to be at the reflection point, hence initial speed ($v_0$) is 0.

Plug in $g$ and solve for $h$:

$$h = 80 + \frac{10 * 3^2}{2} = 80 + 5 * 9 = 80 + 45 = 125 \, (m)$$

As far as we know $h$, we can get $v$ from the conservation law:

$$mgh = \frac{mv^2}{2}$$ $$v^2 = 2gh$$ $$v = \sqrt{2gh}$$ $$v = \sqrt{2 * 10 * 125} = \sqrt{2 * 2 * 5 * 5^3} = \sqrt{2^2 * 5^4} = 2 * 5^2 = 50 \left(\frac{m}{s}\right)$$ $$u \equiv v = 50 \frac{m}{s}$$

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