Question

I need help solving the following question:A grasshopper jumps a horizontal distance of 1.00 m from rest,with an initial velocity at 45.0 degree angle with respect to thehorizontal.Find (a) theinitial speed of the grasshopper and(b) the maximum height reached.

Answer 1

Let's find range x first in terms of horizontalV_x and vertical V_ycomponents of V.

x=V_xt
t= V_y/g
since θ=45 V_x = V_ythen

x= V_x (V_y/g) = V_x²/g
and since V_x = Vcos(45)

x= (Vcos(45))² / g
finally

V= √(gx) / cos(45)

V= √(9.81 x 1.00) / cos(45)

V= 4.43 m/s
Please let me know if you have any questions.

Answer 2

The horizontal distance,Range R=u²sin2θ/g1=u²sin2*45/gFrom this we get the initial speed.The maximum height reachedH=u²sin²θ/2gPlug in the values to get the right solution."Hope this helps!Best of luck for the rest of yourcoursework."