Sagot :
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)
.
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)
.
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.
