
A grasshopper can jump a maximum distance of 1.6 m. It spends negligible time on the ground. How far can it go in 10 s?
Answer
145.8k+ views
Hint: The motion of the grasshopper is like a projectile motion of the object. The maximum distance jumped by the grasshopper is the same as the range. We use the equation for the maximum range of the grasshopper to find the initial speed of the grasshopper.
Complete step by step answer:
Given: The range of the grasshopper is $R = 1.6\;{\text{m}}$and the time of flight of grasshopper is $t = 10\;{\text{s}}$.
Use the equation of the maximum range in the projectile motion to find the initial speed of the grasshopper. The maximum range occurs corresponding to the angle $\theta = 45^\circ $.
$R = \dfrac{{{u^2}}}{g}......\left( 1 \right)$
Here, g is the gravitational acceleration and the value of gravitational acceleration is $10\;{\text{m}}/{{\text{s}}^2}$.
Substitute $R = 1.6\;{\text{m}}$and $g = 10\;{\text{m}}/{{\text{s}}^2}$ in the equation (1) to find the initial speed of the grasshopper.
$1.6\;{\text{m}} = \dfrac{{{u^2}}}{{10\;{\text{m}}/{{\text{s}}^2}}}$
$u = 4\;{\text{m}}/{\text{s}}$
The equation to calculate the distance traveled by the grasshopper is,
$x = ut\cos \theta ......\left( 2 \right)$
Substitute $\theta = 45^\circ $, $u = 4\;{\text{m}}/{\text{s}}$ and $t = 10\;{\text{s}}$in the equation (2) to find distance traveled by grasshopper.
$x = \left( {4\;{\text{m}}/{\text{s}}} \right)\left( {10\;{\text{s}}} \right)\left( {\cos 45^\circ } \right)$
$x = 20\sqrt 2 \;{\text{m}}$
Therefore, the grasshopper can go $20\sqrt 2 \;{\text{m}}$ in the 10 s.
Additional information: The grasshopper remains under the effect of the gravity during the jump. The initial speed of the grasshopper first decreases then increases after reaching to the zero. The maximum distance depends on the angle of the grasshopper.
Note: Be careful in applying the formula of range, as it is nearly the same as the maximum height of the grasshopper. The grasshopper follows the parabolic path because the motion of the grasshopper is the same as the projectile motion. The maximum height attained by the grasshopper depends on the initial speed and angle at which it jumps.
Complete step by step answer:
Given: The range of the grasshopper is $R = 1.6\;{\text{m}}$and the time of flight of grasshopper is $t = 10\;{\text{s}}$.
Use the equation of the maximum range in the projectile motion to find the initial speed of the grasshopper. The maximum range occurs corresponding to the angle $\theta = 45^\circ $.
$R = \dfrac{{{u^2}}}{g}......\left( 1 \right)$
Here, g is the gravitational acceleration and the value of gravitational acceleration is $10\;{\text{m}}/{{\text{s}}^2}$.
Substitute $R = 1.6\;{\text{m}}$and $g = 10\;{\text{m}}/{{\text{s}}^2}$ in the equation (1) to find the initial speed of the grasshopper.
$1.6\;{\text{m}} = \dfrac{{{u^2}}}{{10\;{\text{m}}/{{\text{s}}^2}}}$
$u = 4\;{\text{m}}/{\text{s}}$
The equation to calculate the distance traveled by the grasshopper is,
$x = ut\cos \theta ......\left( 2 \right)$
Substitute $\theta = 45^\circ $, $u = 4\;{\text{m}}/{\text{s}}$ and $t = 10\;{\text{s}}$in the equation (2) to find distance traveled by grasshopper.
$x = \left( {4\;{\text{m}}/{\text{s}}} \right)\left( {10\;{\text{s}}} \right)\left( {\cos 45^\circ } \right)$
$x = 20\sqrt 2 \;{\text{m}}$
Therefore, the grasshopper can go $20\sqrt 2 \;{\text{m}}$ in the 10 s.
Additional information: The grasshopper remains under the effect of the gravity during the jump. The initial speed of the grasshopper first decreases then increases after reaching to the zero. The maximum distance depends on the angle of the grasshopper.
Note: Be careful in applying the formula of range, as it is nearly the same as the maximum height of the grasshopper. The grasshopper follows the parabolic path because the motion of the grasshopper is the same as the projectile motion. The maximum height attained by the grasshopper depends on the initial speed and angle at which it jumps.
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
