What is the maximum height above ground you can achieve with a projectile of mass 0.65 kg, launched from groun

Question

What is the maximum height above ground you can achieve with a projectile of mass 0.65 kg, launched from ground level, if you are able to give it an initial speed of 78.7 m/s?

Answer 1

We know that the maximum height will be reached if you shoot it straight into the air.

vi=78.7 m/s
vf=0
a=-9.8

We know that final velocity (vf) will be 0 because when a projectile reaches it's peak, it changes direction, meaning that at that point, the velocity must be 0. Acceleration=negative gravity (-9.8)

We are trying to find d (displacement). Using the five equations of motion, we know that

vf^2-vi^2=2ad
0^2-(78.7)^2=2(-9.8)(d)
Solve for d.
d=316 m

Actually, there is another way that we can do this using energy. We know that

Ki+ui=Kf+uf

Initial Kinetic+Initial Potential=Final Kinetic+Final Potential

K=(1/2)mv^2 (.5*mass*velocity squared)
u=mgh (mass*gravity*height)

(1/2)mv^2+0=0+mgh

We know that initial potential is 0 because h (height)=0. We know that final kinetic is 0 because v=0 at the max height.

Now, plug in our numbers.

(1/2)(.65)(-78.7)^2=(.65)(-9.8)(h)

h=316 m

I hope that this helps.

Answer 2

First, rockets are no longer projectiles. Projectiles are no longer self propelled; they're projected by skill of different skill. hence the term, projectile. 2d, via fact the projectile is acted on by using skill of the vertical stress of gravity (discounting drag), the time of flight T is set totally by skill of its vertical action over the years. we can write that as: y(t) = h + Uyt - a million/2 gt^2; the place h is the launch top above the end result top, Uy = U sin(theta) is the vertical preliminary velocity = 30 sin(50) and U = 30 mps, g is g, and t = T is the time of flight you're searching for for. you ought to understand the physics of the y(t) equation. the 1st term is the commencing top at t = 0. the 2d term is the top the projectile might circulate if there have been no gravity. The final term is how gravity impacts the vertical flight of the projectile (e.g., pulling it lower back off). via fact the end result's theory to be floor point, we set y(T) = 0 = h + (U sin(theta))T - a million/2 gT^2 and then sparkling up for T given the theory h = 0 additionally. So we've U sin(theta) = a million/2 gT and T = 2U sin(theta)/g = 2*30*sin(50)/9.80 one = 4.sixty 9 sec ANS.

Answer 3

max height = v^2/2*g=78.7^2/(2*9.8)=316meters

Answer 4

james... doesn't his force play a role?