A tier leaps horizontally from a 5.9 meter high rock with a speed of 3.7 m/s
Question: How far from the base of the rock will the tiger land?
See previous post.
2.23
To find the distance from the base of the rock where the tiger will land, we can use the kinematic equation for horizontal motion:
d = v_x * t
Where:
d is the distance from the base of the rock where the tiger will land.
v_x is the horizontal component of the tiger's velocity.
t is the time taken to reach the ground.
Since the tiger is leaping horizontally, the horizontal component of its velocity remains constant throughout the motion. Therefore, we can find v_x by multiplying the initial horizontal velocity (3.7 m/s) by cosθ, where θ is the angle at which the tiger leaps (which in this case is 0 degrees since it is a horizontal leap).
v_x = 3.7 m/s * cos(0°)
v_x = 3.7 m/s
Next, we need to find the time it takes for the tiger to reach the ground. We can use the kinematic equation for vertical motion:
d_y = v_y * t - (1/2) * g * t^2
Where:
d_y is the vertical distance traveled (which is the height of the rock in this case, since the tiger starts at the top).
v_y is the vertical component of the tiger's velocity (which is 0 m/s at the highest point of the trajectory).
g is the acceleration due to gravity (which is approximately 9.8 m/s^2).
t is the time taken to reach the ground.
Since the tiger starts at rest vertically and lands on the ground (d_y = 0), the equation becomes:
0 = - (1/2) * 9.8 m/s^2 * t^2
Simplifying this equation, we find:
4.9 * t^2 = 0
Since t^2 = 0, the only possible value for t is 0. Therefore, the time taken to reach the ground is 0 seconds.
Now, using the equation d = v_x * t, we can calculate the distance from the base of the rock where the tiger will land:
d = 3.7 m/s * 0 s
d = 0 meters
Therefore, the tiger will land at a distance of 0 meters from the base of the rock.