With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 80 m?

Answer 1

u=?,v=0,a=-10m/s^2, S=80m
Use 2aS=v^2-u^2
2*(-10)S=0-u^2
u^2=2*10*80
u^2=1600
u=40m/s. ANS.

Answer 2

Ignoring air resistance, this is exactly the same as asking "If a ball is dropped from 80m, how fast will it be going as it hits the ground?" (which is exactly what happens after the ball begins to drop again to complete its trip in the air.

Ballistic equation (initial velocity=0) is
x = 0.5*a*t^2 where a = g = 9.81m/(s^2) and x = 80m

80m = 0.5*9.81 * t^2
solving for t: t=4.04 seconds

V(t) = a * t = g* t = 9.81m/s^2 * t
V(4.04sec) = 39.84 m/s (meters per second)

Answer 3

Assuming there's no air resistance, you need to throw it at a speed such that its initial kinetic energy exactly matches the final potential energy of the ball at a height of 80 m.

Kinetic energy = Potential energy
KE = PE
1/2*m*v^2 = m*g*h
1/2*v^2 = g*h
v^2 = 2*g*h
v = sqrt(2*g*h)
v = sqrt(2*(9.81 m/s^2)*(80 m))
v = 39.62 m/s

You need to throw the ball with an initial vertical velocity of 39.62 m/s.

Answer 4

human beings make it so perplexing. Its effortless. Kinetic capability at launch = ability capability at max top. So a million/2 mv^2 = mgh m cancels - so which you do not could desire to be responsive to it in any respect v = sqrt (2gh) = sqrt (2.10.eighty) = sqrt 1600 = 40 m/s

Answer 5

kinetic energy given= potential energy of 80 metre
1\2mv2=m*g(10)*80
v=1600root=40m\s

Answer 6

Maximum height: v=at

a=gravity=9.8m/s^2

Distance formula:
d=vt-0.5at^2
80=[(9.8)*t]*t-0.5*9.8*t^2
80=9.8t^2-4.9t^2
80=4.9t^2
t=sqrt(80/4.9)
=4.04seconds

v=9.8*4.04
=39.592 m/s

Answer 7

i can guess the answer
40m/sec because v(square)=2gh where g=10m/sec square.
i hope it shd be correct.