projectile motion question?

Question

The best leaper in the animal kingdom is the puma, which can jump to a height of 4 m when leaving the ground at an angle of 45 degrees.
a) with what speed must the puma leave the ground to reach that height?
b) How far does the puma jump horizontally?

for part a I got 12.62 for the speed.

I am not sure how to solve part b...what equation should I be using? and is the acceleration supposed to be zero for the x component? thank you

Answer 1

What goes up must come down. It takes the same amount of time at it takes to come down assuming there are only forces of gravity acting upon that body.

a) first V=V0 - gt
V= 0 at the very top of 4 m.
V0=gt
So if it has jumped up 4 m (it also then fell 4 m). since
h=0.5gt^2
t=sqrt(2h/g)
finally
V0=g sqrt(2h/g) or if you like
V0=sqrt(2hg)

b) Since it jumped on an angle of 45
V0=Vv (vertical speed component)
Then the horizontal component is
Vh=Vv/tan(45)=Vv

Lets double the time since it takes time to go up and the same amount of time to come down.

t=2(sqrt(2h/g))
finally the horizontal distance covered is
S= Vh t=
S=sqrt(2hg) 2sqrt(2h/g)
S= 2sqrt(4 h^2 )= 4h
S= 16 m

Answer 2

The components of the initial velocity Vi are
Vix = Vi cos 45 = 0.71 V1
Viy = Vi sin 45 = 0.71 V1
a) At the maximum height, Vf is 0. Use Viy
Vf**2 = Vi**2 + 2 gd
0 = (0.71Vi)**2 + 2 (-9.8) (4)
(0.50Vi)**2 = 78.4
Vi = 12.52 m/s
This is the initial speed.

b) still at the maximum height, use Viy
Vf = Viy + gt
0 = 0.71 (12.52) + (-9.8) t
t = 0.91 s
This is the time to reach the maximum height.
If you double the time, this will equal the total time of flight.
For range, use R = Vix t
where t is total time, t = (0.91) 2 = 1.82 s
R = 0.71 (12.52) (1.82) = 16.18 meters
The puma lands 16.18 m away.