Question

If an object is projected horizontally from a height of 5 m with an initial velocity of 7 m/s, what is the value of x^0?
x0=−9.8 m/s2
x subscript 0 baseline equals negative 9.8 m per s squared
x0=0
x subscript 0 baseline equals 0
x0=5
x subscript 0 baseline equals 5
x0=7

Answer 1

Answer:

The value of $x_{0}$ is 5 m.

Explanation:

Given:

The horizontal (initial) velocity of the projection is $u_{h} =7$ m/s.

Height of the object is $h=5$ m.

The vertical (initial) velocity of the projection is $u_{v}=0$ m/s.

The vertical acceleration will be $a_{v}=g=9.8$ $\[\mathrm{m/s^2}\]$

The horizontal acceleration will be $a_{h}=0$ $\[\mathrm{m/s^2}\]$

Now, the time of flight or motion can be calculated as,

$h=u_{v}t+\frac{1}{2}a_{v}t^{2}\\5=0+\frac{1}{2}(9.8)t^{2}\\t=1.01 \mathrm{s}$

Now, the horizontal distance (Range) of the motion $x_{0}$ will be,

$x_{0}=u_{h}t\\x_{0}=5\times1.01\\x_{0}=5.05\\x_{0}\approx 5 \rm{m}$

therefore, the value of $x_{0}$ is 5 m.

For more details, refer he link:

https://brainly.com/question/20689870?referrer=searchResults

Answer 2

The horizontal displacement (x₀) of the object is 7 m.

The given parameters;

• height of the object projection, h = 5 m
• initial horizontal velocity of the object, uₓ = 7 m/s

The vertical displacement equation of the object is given as;

h = vt + ¹/₂gt²

where;

v is the initial vertical velocity = 0

The time of motion is calculated as;

5 = 0  + (0.5 x 9.8)t²

5 = 4.9t²

$t^2 = \frac{5}{4.9} \\\\t^2 = 1.02\\\\t = \sqrt{1.02} \\\\t = 1 \ s$

The horizontal displacement of the object is calculated as;

X₀ = uₓt

X₀ = 7 x 1

X₀ = 7 m

Thus, the horizontal displacement (x₀) of the object is 7 m.

Learn more here:https://brainly.com/question/24527971