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
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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.
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