Respuesta :
Answer:
a) On the distant planet, the ball will travel 395 m horizontally.
b) On the distant planet, the max-height will be 48.3 m.
Explanation:
The equation for the position of an object moving in a parabolic trajectory is the following:
r = (x0 + v0 • t • cos θ, y0 + v0 • t • sin θ + 1/2 • g • t²)
Where:
r = vector position at time t
x0 = initial horizontal position
v0 = initial velocity
t = time
θ = launching angle
y0 = initial vertical position
g = acceleration due to gravity
a) First, let´s calculate the horizontal distance traveled as if we would be on earth. For this, we have to find the magnitude of the vector "r" in the figure. Seeing the figure, we know that the y-component of the vector "r" is 0 if we place the center of the frame of reference at the launching point.
Then, using the equation for the y-component of "r":
y = y0 + v0 • t • sin θ + 1/2 • g • t² (y0 = 0)
0 m = 44.4 m/s · t · sin 26° - 1/2 · 9.8 m/s² · t²
4.9 m/s² · t² = 44.4 m/s · t · sin 26°
t = 44.4 m/s · sin 26° / 4.9 m/s²
t = 3.97 s
Now, we can calculate the x-component of the vector position at final time:
x = x0 + v0 • t • cos θ (x0 = 0)
x = 44.4 m/s · 3.97 s · cos 26° = 158 m
On the distant planet, the ball will travel 158 m · 2.50 = 395 m horizontally.
b) Since the trajectory is parabolic, the maximum height will be reached at the half of the flight. Then, the maximum height will be at t = 3.97 s/2 = 1.99 s.
The max-height on earth wuould be:
y = y0 + v0 • t • sin θ + 1/2 • g • t² (y0 = 0)
y = 44.4 m/s · 1.99 s · sin 26° - 1/2 · 9.8 m/s² · (1.99 s)²
y = 19.3 m
On the distant planet the max-height will be 19.3 m · 2.50 = 48.3 m
Maximum height is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion. The maximum height achieved by the body will be 48.3 meter
The horizontal distance is covered by the body when the body is thrown at some angle is known as the range of the projectile. The range of the ball will be 395 meters.
what is the maximum height achieved in projectile motion?
it is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion. the maximum height of motion is given by
[tex]H = \frac{u^{2}sin^{2}\theta }{2g}[/tex]
[tex]y = y_0+ v_0tsin\theta+\frac{1}{2} gt^{2}[/tex]
[tex]y = 0+ 44.4tsin26^0-\frac{1}{2} \times9.81t^{2}[/tex]
[tex]0 = 0+ 44.4tsin26^0-\frac{1}{2} \times9.81t^{2}[/tex]
[tex]t = \frac{44.4sin26^0}{4.9}[/tex]
[tex]x= x_0+v_0tcos\theta[/tex]
[tex]x= 0+3.97tcos26^0[/tex]
On the distant planet, the ball will travel 2.50 distance traveled on the earth = [tex]2.50\times 158[/tex] = 395 m horizontally
As we know that the maximum height is half of the flight.
so for finding maximum height the time will be t = [tex]\frac{3.975}{2} = 1.995[/tex]
[tex]y = y_0+ v_0tsin\theta+\frac{1}{2} gt^{2}[/tex]
[tex]y = 0+ 44.4\times1.99sin26^0-\frac{1}{2} 9.81\times(1.99)^{2}[/tex][tex]y = 0+ 44.4\times1.99sin26^0-\frac{1}{2} 9.81\times(1.99)^{2}[/tex]
y = 193 meter.
Hence the range of the ball will be 19.3 meters.
On the distant planet, the ball will travel 2.50 distance traveled on the earth = [tex]19.3\times2.50[/tex] = 48.3 m horizontally
To learn more about the range of projectile refer to the link ;
https://brainly.com/question/139913
