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.

Answer 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

Answer 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.

Answer 3

correct