Question

A projectile is launched with an initial speed of 42 feet per second. It is projected at an angle of 50 degrees. How far does the projectile travel? How much farther does it travel if it is launched with an initial speed of 84 feet per second?

Answer 1

Answer:

1) 53.95 ft ( approx )

2) 161.85 ft. ( approx )

Step-by-step explanation:

Given,

Initial velocity, $v_0=42\text{ ft per second}$

Also, the projected angle, $\theta=50^{\circ}$

1) Thus, the distance travelled horizontally ( Range of the projectile motion ),

$R=\frac{v_0^2sin2\theta}{g}$

Where, g is the acceleration due to gravity that having the value 32.2 ft per second,

$=\frac{42^2 sin 100^{\circ}}{32.2}$

$=53.9503377737$

$\approx 53.95$

Hence, the projectile covers 53.95 feet ( approx )

2) If $v_0=84\text{ ft per sec}$

Then the range would be,

$R=\frac{84^2 sin 100^{\circ}}{32.2}$

$=215.801351095$

$\approx 215.80$

Thus, the additional distance = 215.80 ft - 53.95 ft = 161.85 ft.

Hence, It will travel 161.85 ft extraa.

Answer 2

Using the formulas for projectile motion, we can find the range of a projectile. Horizontal range of a projectile is (Vo^2(sin2angle))/g. For an initial velocity, Vo=42 ft/s and angle of 50 degrees, the horizontal range of the projectile is 53.95 ft. If the initial velocity is changed to Vo=84 ft/s, the horizontal range is 215.80 ft. This is 161. 85 feet farther from the first projectile.