College Physics by Openstax Chapter 3 Problem 25: Computing Horizontal and Vertical Displacement of a Projectile in Two-Dimensional Motion

Problem:

A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

Solution:

Since we do not know the exact location of the projectile after 3 seconds, consider the following arbitrary figure:

The path of the projectile from the ground to a point 3 seconds later.

From the figure, we can solve for the components of the initial velocity.

\begin{align*} \text{v}_{\text{ox}} &=\left(50\:\text{m/s}\right)\cos 30^{\circ} \\ & =43.3013\:\text{m/s} \\ \\ \text{v}_{\text{oy}} & =\left(50\:\text{m/s}\right)\sin 30^{\circ} \\ &=25\:\text{m/s} \\ \end{align*}

So, we are asked to solve for the values of x and y. To solve for the value of the horizontal displacement, x, we shall use the formula x=v_oxt. That is,

\begin{align*} \text{x} & =\text{v}_{\text{ox}}\text{t} \\ \text{x} & =\left(43.3013\:\text{m/s}\right)\left(3\:\text{s}\right) \\ \text{x} & =130\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

To solve for the vertical displacement, y, we shall use the formula y=v_oyt+1/2at². That is

\begin{align*} \text{y} & =\text{v}_{\text{oy}}\text{t}+\frac{1}{2}\text{a}\text{t}^2 \\ \text{y} & =\left(25\:\text{m/s}\right)\left(3\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(3\:\text{s}\right)^2 \\ \text{y} & =30.9\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Therefore, the projectile strikes a target at a distance 129.9 meters horizontally and 30.9 meters vertically from the launching point.

College Physics Chapter 3 Problems

Problem 1	Problem 2	Problem 3	Problem 4	Problem 5
Problem 6	Problem 7	Problem 8	Problem 9	Problem 10
Problem 11	Problem 12	Problem 13	Problem 14	Problem 15
Problem 16	Problem 17	Problem 18	Problem 19	Problem 20
Problem 21	Problem 22	Problem 23	Problem 24	Problem 25
Problem 26	Problem 27	Problem 28	Problem 29	Problem 30
Problem 31	Problem 32	Problem 33	Problem 34	Problem 35
Problem 36	Problem 37	Problem 38	Problem 39	Problem 40
Problem 41	Problem 42	Problem 43	Problem 44	Problem 45
Problem 46	Problem 47	Problem 48	Problem 49	Problem 50
Problem 51	Problem 52	Problem 53	Problem 54	Problem 55
Problem 56	Problem 57	Problem 58	Problem 59	Problem 60
Problem 61	Problem 62	Problem 63	Problem 64	Problem 65
Problem 66	Problem 67	Problem 68	Problem 69	Problem 70