2 physicss questions?

Question

A squirrel jumps from one branch to another on 3 meters away at the same height. If hhe jumps at an angle of 20 degrees what minimum speed must he have in order to make the jump?

A baseball is thrown vertically upward with a speed of 25 m/s from a height of 2 meters. How long does it take it to reach its heighest point? How long does it take the ball to hit the ground after it reaches its highest point?

Answer 1

baseball one -> use formula t= (vf-vi)/a (0-25)/-9.81
takes 2.55 seconds to reach highest point

the highest pt is d= (vf squared - vi squared) / 2a 625/(2)(9.81)
the highest pt reached is 31.855 so add 2 to get to the ground... 33.855

t = square root ((2d)/a) square root ((2*33.855)/9.81)
therefore the time it takes for it to get back to the ground is 2.62 seconds

Answer 2

Let him start with speed u. The horizontal component is uCos20 and the vertical component is uSin20.

Travelling horizontally he covers 3m at a speed of uCos20. The time taken is t = 3/(uCos20)

The time taken to travel to the maximum height can be determined from v = u + a*t, where a = -g, and we use vertical speeds.

From this we get t = vSin20/g, thus the time taken to travel up and down is 2vSin20/g

These two times must be equal giving us

3/(uCos20) = (2uSin20)/g. Solve for u.

For the second question, calculate the time taken to reach max height with an initial velocity of 25 m/s.

v = u + at
0 = 25 -(9.8)t
t = 25/9.8 = 2.6s

It will then take 5.2 seconds to return to it's starting position, only this time it has a downwards velocity of 25 m/s.

Calculate how long it takes to travel 2 m, with an initial velocity of 25 m/s.

s = ut + (1/2)*a*t^2. Solve this and add to the 5.2 s determined earlier

Answer 3

actually g doesn't depend on mass, and so as long as we're treating the squirrel as a reasonably sized object it could be 5 lbs or 50000000000 lbs.
So, as for your question:
1) use the range equaiton! Otherwise you will need to do a trig substitution.
D = v^2sin(2theta)/(g)
D*g /(sin2theta) = v^2
3 * 9.80/(sin 44) = v^2
square root and solve!
2) set up the knowns:
y:
v0 = 25 m/s
a = -9.8 m/s^2
v = 0
t = ?
now use a = (final v + initial v)/(2t) to solve
For the second part:
same knowns, and x = -2.0m. Then solve using:
x = v0t + (1/2) a t^2

Answer 4

to squirrel part 6.763m/s
to baseball part 2.55s, 2.6298s
explanation:when squirl lands he will be on same hight as when he jumped so use formula Sy=ut + 1/2at^2
Sy=0 ,u=usin20, a=9.8, t=t
so 0=usin20t + 1/2(9.8)t^2
0=t(usin20 + 4.9t)
so t=0 (start) or t=.0698u
so now use formula Sx=ut + 1/2at^2
Sx=3, u=ucos20, a=0, t=.0698u
so 3=ucos20(.0698u) + 0
3=.06559u^2
u^2=45.7
u=6.763m/s

to baseball part
at highest point when velocity is 0
use formula v=u+at
v=0, u=25, a=-9.8, t=t
so 0=25-9.8t
t=2.55s

to answer 2nd part need to find how far ball travelled upwards
s=ut+1/2at^2
s=s, u=25, a=-9.8, t=2.55
so s=25*2.55 -1/2*9.8*6.5
so s=31.89m
now use same formula again
s=31.89+2, u=0, a=9.8, t=t
so 33.89=0 + 1/2*9.8*t^2
33.89=4.9t^2
t^2=6.9
t=2.6299s

Answer 5

No, the airplane isn't travelling quicker than sound, however the postpone you hear is from sound moving slower than the cost at which the airplane is recognizable to the attention. Sound travels at 350 meters in line with 2d in customary air, no longer counting decreased density on the top of a airplane. let's imagine your airplane is 30,000 ft severe (9144 meters), it would take sound 26.one million 2d to realize floor.

Answer 6

Vh = Vcos20
Vv = Vsin20
0 - Vv = -at
t = 2Vv/a
d = 2V^2sin20cos20/a
V^2 = (9.8)(3)/(2sin20cos20)
V = 6.763 m/s

t1 = 25/9.8 = 2.551 s

0 = - 2 + (1/2)(9.8)t^2 - 25t
t = (25 ± √(625 + 39.2))/9.8 - 2.551
t = (25 ± √664.2)/9.8 - 2.551
t = (25 ± 25.772)/9.8 - 2.551
t = 2.6298 s

Answer 7

there is missing the weight of the squirl and the basebal