To determine how far away the frog lands, you can use the principles of projectile motion. Here’s a step-by-step approach to solve for the horizontal distance:
1. **Resolve the velocity into horizontal and vertical components**:
\[
v_{x} = v \cos(\theta)
\]
\[
v_{y} = v \sin(\theta)
\]
where \( v = 8.40 \, \text{m/s} \) and \( \theta = 58.7^\circ \).
Let's calculate these:
\[
v_{x} = 8.40 \cos(58.7^\circ)
\]
\[
v_{y} = 8.40 \sin(58.7^\circ)
\]
Using the trigonometric values,
\[
\cos(58.7^\circ) \approx 0.5261 \quad \text{and} \quad \sin(58.7^\circ) \approx 0.8508
\]
Therefore,
\[
v_{x} = 8.40 \times 0.5261 \approx 4.42 \, \text{m/s}
\]
\[
v_{y} = 8.40 \times 0.8508 \approx 7.14 \, \text{m/s}
\]
2. **Determine the time of flight**:
The time of flight \( t \) can be determined by analyzing the vertical motion. The frog will take the same amount of time to go up and come down under gravity.
The total time of flight \( t \) is calculated from:
\[
t = \frac{2 v_{y}}{g}
\]
where \( g \approx 9.81 \, \text{m/s}^2 \).
Substituting \( v_{y} \approx 7.14 \, \text{m/s} \):
\[
t = \frac{2 \times 7.14}{9.81} \approx \frac{14.28}{9.81} \approx 1.46 \, \text{s}
\]
3. **Calculate the horizontal distance**:
The horizontal distance \( x \) (also known as range) can be calculated by using the horizontal component of the velocity and the time of flight:
\[
x = v_{x} \times t
\]
Substituting \( v_{x} \approx 4.42 \, \text{m/s} \) and \( t \approx 1.46 \text{s} \):
\[
x = 4.42 \times 1.46 \approx 6.45 \, \text{m}
\]
Therefore, the frog lands approximately **6.45 meters** away.
a frog leaps at 8.40 m/s at a 58.7 degrees angle to the horizon. HOw far away does the frog land?
trianglex=[?]m
1 answer