Answered

A projectile is launched from ground level to the top of a cliff which is 195m away and 155m high. If the projectile lands on top of the cliff 7.6s after it is fired, find the initial velocity of the projectile (mag. and direction). Neglect air resistance.



Answer :

Refer to the figure shown below.

u = initial launch velocity, m/s
θ = launch angle, north of east, deg.

Wind resistance is ignored.
g = 9.8 m/s².

The horizontal component of the launch velocity is u cosθ.
Because the horizontal distance traveled in 7.6 s is 195 m, therefore
7.6*(u cosθ) = 195
u cosθ = 25.658                (1)

The vertical component of the launch velocity is u sinθ.
Because the vertical height traveled in 7.6 s is 155 m, therefore
(u sinθ)*7.6 - 0.5*9.8*(7.6²) = 155
u sin θ - 37.24 = 20.395
usin θ = 57.635               (2)

From (1) and (2), obtain
tan θ = 57.635/25.658 = 2.246
θ = 66°

From (2), obtain
u = 57.635/sin(66°) = 63.09 m/s

Answer:
Velocity  = 63.09 m/s
Direction = 66° north of east  (or 66° measured counterclockwise from the x-axis)
.


View image Аноним

The initial velocity with which the projectile is launched is [tex]\fbox{\begin\\63.08\text{ m/s}\end{minispace}}[/tex] and the projectile is fired at [tex]\fbox{\begin\\65.99^\circ\end{minispace}}[/tex] with repect to ground.

Further Explanation:

The velocity of the projectile is defined as the distance covered by the projectile per unit time.

Given:

The horizontal distance between the initial position of projectile and the cliff is [tex]195\text{ m}[/tex].

The height of the cliff is [tex]155\text{ m}[/tex].

The time interval after which the projectile land on the top of the cliff is [tex]7.6\text{ s}[/tex].

Concept:

The projectile is fired in air, in front of a cliff of height [tex]155\text{ m}[/tex]. The projectile will have horizontal and vertical components of its initial velocity.

The horizontal component of the velocity is:

[tex]\fbox{\begin\\u_x=\dfrac{x}{t}\end{minispace}}[/tex]                                  ...... (1)

Here, [tex]u_x[/tex] is the horizontal component of velocity, [tex]x[/tex] is the horizontal distance and [tex]t[/tex] is the time interval after which the projectile land on the top of the cliff.

The vertical component of velocity is:

[tex]\fbox{\begin\\u_y=\dfrac{h}{t}+\dfrac{gt}{2}\end{minispace}}[/tex]                         ....... (2)

Here, [tex]u_y[/tex] is the vertical component of velocity, [tex]h[/tex] is the height of the cliff or the vertical distance covered by the projectile and [tex]g[/tex] is the acceleration due to gravity.

The initial velocity with which the projectile is fired is:

[tex]\fbox{\begin\\u=\sqrt{{u_x}^{2}+{u_y}^{2}}\end{minispace}}[/tex]                                ...... (3)

Here, [tex]u[/tex] is the initial velocity with which the projectile is fired.

The direction in which the projectile is fired is:

[tex]\fbox{\begin\\\theta =tan^{-1}\dfrac{u_y}{u_x}\end{minispace}}[/tex]                            

   ...... (4)

Here, [tex]\theta[/tex] is the direction in which the projectile is fired with respect to ground.

Substitute the values in equation (1).

[tex]\begin{aligned}u_x&=\dfrac{195\text{ m}}{7.6\text{ s}}\\&=25.66\text{ m/s}\end{aligned}[/tex]

Substitute the values in equation (2).

[tex]\begin{aligned}u_y&=\dfrac{155\text{ m}}{7.6\text{ s}}+\dfrac{(9.8\text{m}/\text{s}^2)(7.6\text{ s})}{2}\\&=57.63\text{ m/s}\end{aligned}[/tex]

Substitute the values in equation (3).

[tex]\begin{aligned}u&=\sqrt{(25.66)^2+(57.63)^2}\\&=63.08\text{ m/s}\end{aligned}[/tex]

Substitute the values in the equation (4).

[tex]\begin{aligned}\theta&=tan^{-1}\dfrac{57.63}{25.66}\\&=65.99^\circ\end{aligned}[/tex]

Thus, the initial velocity with which the projectile is launched is [tex]\fbox{\begin\\63.08\text{ m/s}\end{minispace}}[/tex] and the projectile is fired at [tex]\fbox{\begin\\65.99^\circ\end{minispace}}[/tex] with repect to ground.

Learn more:

1.  The motion of a body under friction brainly.com/question/4033012

2.  A ball falling under the acceleration due to gravity brainly.com/question/10934170

3. Conservation of energy brainly.com/question/3943029

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Projectile, projectile is fired, cliff, vertical velocity, horizontal velocity, direction, magnitude of velocity, 63.08m/s, 63m/s, 63.1 m/s, 65.99 degree, 66 degree, 65.9 degree.

View image avantikar