physics

A grasshopper jumps a horizontal distance of 1m from rest, with an initial velocity at a 45 degree angle with respect to the horizontal. Find the initial speed of the grasshopper and the maximum height reached.

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  1. vy=vsin45
    vx=vcos45

    hf=hi-1/2 g t^2+vsin45*t
    t=2vsin45/g in the air.
    distance horizontal=vcos45*t
    1m=v .707*2v*.707/g
    1=v^2/2g
    v=sqrt19.8 m/s
    max height:
    mgh=1/2 m v^2
    h=1/2 (19.8)/g

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    bobpursley
  2. we know for sure that a projectile reaches it's maximum horizontal distance when it makes an angle of 45° with the horizontal. from this we can find a relation between the distance and velocity which is d=v^2/g where g is a due to gravity,d displacement in the horizontal and v the initial velocity of launch. plug in the given values in tho equation d=^2/g(1=v^2/9.8) making v the subject gives v=√(1*9.8) which gives a initial velocity of 3.31.give the initial velocity u can fund the maximum height the grasshopper jumped in the vertical direction.

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  3. correct

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