a grasshopper can jump a maximum horizontal distance of 40cm.if it spends negligible time on the ground then in this case its speed along the horizontal road will be???
options are:-
a...1m/s
b...2m/s
c...underoot 2m/s
d...2underroot2m/s
Answers
Answered by
7
Considering that the grasshopper starts on the ground and lands on the ground, there is no change in elevation. The range of the grasshopper is given by: \[R=\frac{v^{2}\sin 2\theta }{g}\] If we now take the information that this is a maximal distance, the maximum range is given when the angle of launch is 45 degrees. This yields the equation: \[R_{max}=\frac{v^{2}}{g}\] We know R_max=0.4 m and g=9.8 m/s^2, so we can solve for v: \[v=1.98 m/s\] We are asked for the horizontal component, which by trigonometry is: \[v_{x}=v\cos\theta \] We again take our angle of launch as 45 degrees and arrive at: \[v_{x}=1.40 m/s \]
Similar questions