mechanics question, really need help :(?

Question

I'm really confused, can someone help me please with this question, or explain how to do it

Pluto the dog jumps between 2 buildings
1) in the first stunt, the buildings are 40m above ground and the gap between them is 6m. If the dog takes off at 8m/s what range of angles to the horizontal must he take off at in order to complete the jump?

2)in the 2nd stunt the gap between the buildings is 8m, but a takeoff ramp is provided. show my differention that if the dogs take off speed is V0, the maximum range occurs for takeoff angles of 45degrees.

3)find the minumum takeoff speed at which the dog can make the 8m jump.

Please help me!

Answer 1

If the dog jumps straight up velocity v0 he will hit the ground with velocity -v0, which requires a total velocity change of 2v0. The amount of time required for this to take place is 2v0/g.
If the dog jumps at angle theta to the horizontal with initial velocity v0 then its vertical velocity component vv will be equal to v0 * sin(theta) and its horiziontal velocity component vh will be v0 * cos(theta).

The dog will hit the ground after time 2vv/g, at which point it will have travelled (2vv/g)*vh horizontally. This works out to 2 * v0^2 * sin(theta)*cos(theta)/g, which simplifies to v0^2/g * sin(2*theta). This is consistent with the fact that a 45 degree jump will travel the farthest distance.

1.)
trig identity: sin2Θ = 2sinΘcosΘ
v(zero horz) = 8 cosΘ
v(zero vert) = 8sinΘ
t = 2v(zero vert)/g = 2(8sinΘ)/g
s(horz) = 8cosΘ(2)(8sinΘ/g
s(horz) = (64/g)sin(2Θ) = 6
sin2Θ = 0.91875
2Θ = 66.744degrees; 2Θ = 180 - 66.744degrees
Θ = 33.372degrees; Θ = 56.028degrees
Since the maximum is at 45 degrees we can conclude successful jumps in this range:
33.372deg
2.)
s(horz) = (64/g)sin2Θ
ds/dΘ = (128/g)cos2Θ = 0
cos2Θ = 0
2Θ = 90
Θ = 45

3.)
s(horz) = (v(zero)cosΘ)(v(zero)sinΘ)/g
8 = v(zero)^2cos45(sin45)/g
8 = v(zero)^2(√2)^2/g
v(zero)^2 = 8g/2
v(zero) = √(4g) = 6.261m/s