Question

A projectile is launched with a velocity of 25 m/s at an angle of 20 degrees with respect to the horizontal. What is its initial velocity along the x-axis?

Select one:
a. 10 m/s
b. 23.49 m/s
c. 18.5 m/s
d. 75 m/s
e. None of the choices give.


A projectile is launched with a velocity of 25 m/s at an angle of 20 degrees with respect to the horizontal. What is its initial velocity along the y-axis?

Select one:
a. 2.09 m.s
b. 4.21 m/s
c. 17.10 m/s
d. 8.55 m/s
e. None of the choices given.

A projectile is launched with a velocity of 25 m/s at an angle of 20 degrees with respect to the horizontal. How long will it take for it to reach the top of its flight?

Select one:
a. 5.21 s
b. 3.28 s
c. 1.74 s
d. .87 s
e. None of the choices given.
A projectile is launched with a velocity of 25 m/s at an angle of 20 degrees with respect to the horizontal. What is its range?



Select one:
a. 40.8 m
b. 81.6 m
c. 33.21 m
d. 20.4
e. None of the choices given.

Answer 1

Answer:

b. 23.49 m/s

d. 8.55 m/s

d. .87 s

a. 40.8 m

Explanation:

The initial along the x-axis is given by the following formula:

$V_{ox} = V_{o}Cos\theta\\\\V_{ox} = (25\ m/s)(Cos\ 20^{o})\\\\V_{ox} = 23.49\ m/s$

Hence, the correct option is:

b. 23.49 m/s

The initial along the x-axis is given by the following formula:

$V_{oy} = V_{o}Sin\theta\\V_{oy} = (25\ m/s)(Sin\ 20^{o})\\V_{oy} = 8.55\ m/s$

Hence, the correct option is:

d. 8.55 m/s

Now, for the time to reach maximum height:

$t = \frac{V_{o}Sin\theta}{g}\\\\t = \frac{(25\ m/s)Sin\ 20^{o}}{9.8\ m/s^{2}}\\\\t = 0.87\ s$

Hence, the correct option is:

d. .87 s

For the range of projectile:

$R = \frac{V_{o}^{2}\ Sin\ 2\theta}{g}\\\\R = \frac{(25\ m/s)^{2}(Sin\ 40^{o})}{9.8\ m/s^{2}}\\\\R = 40.9\ m$

Hence, the closest option is:

a. 40.8 m