Question

A ball is kicked such that it leaves the ground at an upwards angle with an initial velocity of vi.
The ball takes 3.1 s to hit the ground 26 m from where it was kicked.
What is the magnitude of the initial velocity of the ball? (in m s−1 to 2.s.f)
(note: ignore the effects of air resistance when solving this problem)

Answer 1

Task:

$t_0=3.1 \mathrm{\ s} \\ L=26 \mathrm{\ m}$

Find: $v_0$.

Solution:

Coordinate on the y-axis can be expressed as:

$y(t)=\int\int a_y dt dt = \int\int -g dt dt = \int(v_{0y} - gt)dt = y_0+ v_{oy}t - \cfrac{gt^2}{2}$

It is known that the initial y-coordinate $y(0)=y_0=0$ m and it will be the same at the time $t_0$, so:

$v_{0y}t-\cfrac{gt_0^2}{2}=0\Rightarrow v_{0y}=\cfrac{gt_0}{2}$

Coordinate on the x-axis can be expressed as:

$x(t)=\int \int a_xdtdt=\int v_{0x} dt=x_0+v_{0x}t$

It is known that the initial x-coordinate $x(0)=x_0=0$ and it will be equal to $L$ at the time $t_0$, so:

$v_{0x}t_0=L\Rightarrow v_{0x}=\cfrac{L}{t_0}$

Thus, according to the Pythagorean theorem the initial velocity $v_0$ is equal to:

$v_0=\sqrt{v_{0x}^2+v_{0y}^2}=\sqrt{(\cfrac{L}{t_0})^2+(\cfrac{gt_0}{2})^2}=\sqrt{\cfrac{L^2}{t_0^2}+\cfrac{g^2t_0^2}{4}}=\sqrt{\cfrac{26^2}{3.1^2}+\cfrac{9.8^23.1^2}{4}}\approx 17.35\mathrm{\ m/s}$

Answer: $v_0\approx17.35 \mathrm{\ m/s}.$