A grasshopper hops along a level road. On each hop, the grasshopper launches itself at angle θ0 = 45ο and achieves a range R = 0.80 m. What is the average horizontal speed of the grasshopper as it hops along the road? Assume that the time spent on the ground between hops in negligible and achieves a range. What is the average horizontal speed of the grasshopper as it hops along the road? Assume that the time spent on the ground between hops is negligible.
The Correct Answer and Explanation is :
To find the average horizontal speed of the grasshopper, we must first determine the time of flight for each hop and the horizontal component of velocity during the hop.
Step 1: Range of a projectile
The range ( R ) of a projectile is given by:
[
R = \frac{v_0^2 \sin 2\theta_0}{g}
]
where:
- ( v_0 ) = initial velocity of the grasshopper
- ( \theta_0 = 45^\circ )
- ( g = 9.8 \, \text{m/s}^2 ) (acceleration due to gravity)
Given ( R = 0.80 \, \text{m} ), substituting ( \sin 2\theta_0 = \sin 90^\circ = 1 ):
[
0.80 = \frac{v_0^2}{9.8}
]
[
v_0^2 = 0.80 \times 9.8 = 7.84
]
[
v_0 = \sqrt{7.84} \approx 2.8 \, \text{m/s}
]
Step 2: Time of flight
The time of flight ( T ) for a projectile is given by:
[
T = \frac{2v_0 \sin \theta_0}{g}
]
Since ( \theta_0 = 45^\circ ) and ( \sin 45^\circ = \frac{\sqrt{2}}{2} ):
[
T = \frac{2(2.8)(\frac{\sqrt{2}}{2})}{9.8} = \frac{2.8 \sqrt{2}}{9.8}
]
[
T \approx 0.4 \, \text{seconds}
]
Step 3: Average horizontal speed
The horizontal velocity of the grasshopper during each hop is:
[
v_x = v_0 \cos \theta_0
]
[
v_x = 2.8 \cdot \cos 45^\circ = 2.8 \cdot \frac{\sqrt{2}}{2} \approx 2.0 \, \text{m/s}
]
Since the time spent on the ground is negligible, the average horizontal speed is equal to the horizontal velocity during each hop:
[
\text{Average horizontal speed} = v_x \approx 2.0 \, \text{m/s}.
]
Explanation (300 words)
The grasshopper’s motion can be analyzed as projectile motion, where it follows a parabolic trajectory. The given information includes the launch angle (( \theta_0 = 45^\circ )) and the range (( R = 0.80 \, \text{m} )). Using the equation for the range of a projectile, we calculated the grasshopper’s initial velocity (( v_0 )). Since the angle of launch is ( 45^\circ ), the range equation simplifies as ( \sin 2\theta_0 = 1 ).
With ( v_0 = 2.8 \, \text{m/s} ), we determined the total time of flight (( T )) using the vertical motion equations. This time corresponds to the time the grasshopper is in the air during one hop. The horizontal component of the grasshopper’s velocity (( v_x )) remains constant throughout the hop because air resistance is neglected.
Finally, since the problem specifies that the time spent on the ground is negligible, the grasshopper’s average horizontal speed is the same as its horizontal velocity during the hop, ( v_x = 2.0 \, \text{m/s} ). This result highlights the utility of decomposing projectile motion into horizontal and vertical components to analyze average speed and trajectory effectively.