A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 9.3 m from the takeoff point. If the kangaroo leaves the ground at a 17.6 degree angle, what is its takeoff speed?
L=vₒ²•sin2α/g,
vₒ =sqrt(L•g/ sin2α)=
=sqrt(9.3•9.8/sin35.2º)=2.26 m/s
To find the takeoff speed of the kangaroo, we need to break down the motion into horizontal and vertical components.
Let's assume the takeoff speed is denoted as "v."
Since the kangaroo leaves the ground at a 17.6 degree angle, we can determine the horizontal component of the velocity, which can be represented by "v_x."
The horizontal velocity is given by v_x = v * cosθ, where θ is the angle of takeoff (17.6 degrees in this case).
Similarly, the vertical component of the velocity can be represented by "v_y."
The vertical velocity is given by v_y = v * sinθ.
In this case, the distance covered horizontally is given as 9.3 m. We can use this information to solve for the horizontal component of the velocity.
9.3 m = v_x * t, where t is the time of flight.
Since the horizontal velocity remains constant throughout the flight, v_x = v * cosθ.
Substituting this into the equation, we have:
9.3 m = v * cosθ * t.
Now, let's consider the vertical component of the motion. We know that the motion is affected by gravity, causing the kangaroo to move in a parabolic trajectory. The vertical displacement of the kangaroo at the highest point is zero.
Using the equation for vertical displacement, we have:
0 = v_y * t - (1/2) * g * t^2,
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values, we have:
0 = v * sinθ * t - (1/2) * 9.8 * t^2.
Now, we have two equations and two unknowns. We can solve them simultaneously.
From the equation for horizontal displacement, we have:
v * cosθ = 9.3 / t.
Simplifying, we get:
v = (9.3 / t) / cosθ.
Substituting this value into the equation for vertical displacement, we have:
0 = ((9.3 / t) / cosθ) * sinθ * t - (1/2) * 9.8 * t^2.
Simplifying this equation gives:
0 = (9.3 / t) * sinθ - 4.9 * t.
Rearranging the equation, we have:
(9.3 / t) * sinθ = 4.9 * t.
Taking the reciprocal of both sides, we get:
t / (9.3 * sinθ) = 1 / (4.9).
Solving for t, we find:
t = (9.3 * sinθ) / (4.9).
Now, we can substitute this value of t back into the equation for v:
v = (9.3 / t) / cosθ.
Substituting the value of t, we get:
v = (9.3 / [(9.3 * sinθ) / (4.9)]) / cosθ.
Simplifying further, we find:
v = (4.9 * cosθ) / sinθ.
Finally, substituting the value of θ (17.6 degrees), we get:
v = (4.9 * cos(17.6°)) / sin(17.6°).
Evaluating this expression gives the takeoff speed of the kangaroo.