A tiger leaps horizontally from a 15.0 m high rock with a speed of 7.10 m/s. How far (in meters) from the base of the rock will it land?
how long does it take to fall 15m?
h=1/2 g t^2 solve for t.
then, distancewenthorizontal: 15*time
is g negative or positive, because that is what i did and the answer is still wrong!
final position is -15. g is -9.8m/s^2.
the negative direction is up to you. You decide on the frame of reference. Most choose up as positive, downwards as negative.
so h is 15, g is 9.8
15= 4.9t^2
then take the square root of 3.061. then do i multiply by 15?
To find out how far the tiger will land from the base of the rock, we can use the principles of projectile motion. We need to determine the horizontal displacement.
First, let's break down the problem.
Given:
- Initial vertical height (h) = 15.0 m
- Horizontal velocity (v) = 7.10 m/s
- Vertical acceleration due to gravity (g) = 9.81 m/s²
To find the horizontal displacement (x), we can use the following equation of motion:
x = v * t
To determine the time of flight (t), we'll use the equation that relates the vertical displacement, initial vertical velocity, and acceleration due to gravity:
h = (1/2) * g * t²
Rearranging the equation, we can solve for t:
t = sqrt((2 * h) / g)
Substituting the given values:
t = sqrt((2 * 15.0) / 9.81)
Evaluate the value of t:
t ≈ 1.96 seconds
Now, we can substitute the value of t into the equation for horizontal displacement:
x = v * t
x = 7.10 m/s * 1.96 s
Evaluate the value of x:
x ≈ 13.916 m
Therefore, the tiger will land approximately 13.916 meters from the base of the rock.