Physicssss help?

Question

a long jumper leaves the ground at an angle of 22 degrees to the horizontal and at a speed of 12.5m/s A) How far does he jump? B) What is the maximum height reached? C) How long is he in the air?


Answers are 11.1m, 1.12m, and 0.955 sec
I have no idea how to get those answers!!! please help

Answer 1

If you analyse the speed v into two
perpendicular directions
Vy (perpend.to the ground) and
Vx (parallel to the ground) you have
Vx=V*cos22=11.59
Vy=V*sin22=4.68
B)For the movement with Vy you have
when he reaches the max height
0=Vy-g*t<-->t=Vy/g<-->t=0.477 sec
And h(max)=Vy*t-0.5*g*t^2=1.12 aprox.
C)obviously he is in the air for
T=2*t=0.955 aprox.
A)For the movement in Vx direction
S=Vx*T=11.59*0.955=11.1 m

Answer 2

Hi,
Well, we know that these numbers aren’t going to be very accurate because of the distributed mass, but so be it.
Part A:
Now, we know that the equation for the range, the distance he jumps, is this:
R = (Vo² sin 2θ)/g (Where g = 9.8 m/s²)
= (12.5² sin 44)/9.8
= 11.07 which when rounded off = 11.1 m
If you work this on your calculator, be sure you have it set for degrees rather than radians.

Part B:
h = vosin² θ/2g
= 12.5 (sin 22)² /(2*9.8)
= 1.11 m

Part C:
t= 2(vo sin θ/g)
= 2((12.5*sin 22)/9.8)
= 2(.4778)
=.9556

You didn’t say if you need to know how to get the equations, and I don’t want to overload you, but here are a few hints in case you need them. Start with this equation:
Vy = Vyo-gt = Vo sin θ –gt
Now, set Vy = o and solve for t.
Substitute this term for t into this equation:
Y = (Vosin θ)t – (1/2)gt² and solve for h.

The range can be derived from this equation:
X = (Vocosθ)t

Hope this helps.
FE