Question

A circus cat has been trained to leap off a 12-m-high platform and land on a pillow. The cat leaps off at v0 = 3.9 m/s and an angle θ = 41° (see figure below). (a) Where should the trainer place the pillow so that the cat lands safely? d = 5.43 Correct: Your answer is correct. m (b) What is the cat's velocity as she lands in the pillow? (Express your answer in vector form.) vf = Incorrect: Your answer is incorrect. m/s

Answer 1

The initial x component of velocity
v_0x = v₀ cos
Where v₀ is the initial velocity with which the cat is projected, is the angle of projection.
v_0x = 3.9 m/s cos 41
v_0x = 2.94 m/s
The initial y component of the velocity
v_0y = v₀ sin
v_0y  = 2.559 m/s
The velocity of the cat at any later times can be written as
v = v₀ + a t
The x component
v_x = v_0x + a_x t
There is no acceleration along the x direction, so the a_x = 0. The x component of velocity at all times is v_0x
v_x = 2.94 m/s
The y component of velocity
v_y = v_0y + a_y t
v_y = 2.559 m/s + g t
The acceleration in the y direction is g.
Next, we see the displacements at any time.
The general relation is
s = v₀ t + (1/2) a t² + s₀
The x direction,
s_x = v_0x t + s₀
The y direction,
s_y = v_0y t + (1/2) g t² + s_0y
We have to find the time t to find the velocity of the cat at the time of impact, At the time of impact s_y = 0
0 = v_0y t + (1/2) g t² + s_0y
0 = 2.559 t - (1/2) x 9.81 x t² + 12
This is a quadratic equation in t, the solutions are
t = 1.847 and t = -1.3254,
We take the positive value, so t = 1.847 s
We can find the velocity at 1.847 s
v_y(1.847s) = 2.559 m/s + (-9.81m/s² x 1.847)
v_y = -15.56 m/s
The x component of velocity is the same throughout the motion
v_x = 2.94 m/s
The total velocity in the component form,
v = v_x i + v_y j^{^}
v = 2.94 m/s i^{^} - 15.56 m/s j^{^}