Maximum height of ball thrown vertically!!?

Question

A ball with a mass of 0.7 kg is thrown vertically into the air and has an initial
velocity of 10 m s-1. Calculate the maximum height that the ball will reach???

Answer 1

initial velocity = 10 m/s

final velocity = 0 m/s....since the ball is thrown vertically there is that point where the ball is no longer accelerating up or down.

acceleration due to gravity = -9.8m/s^2

just use 1 of the four kinematioc equations and plug in ur knowns and solve. I dont believe you will need the mass for this problem so I think u can ignore that. I dont have the eqs in front of me or I would do more, sorry. But I hope Ive helped. Good luck

Answer 2

Physics? ball thrown upward vertically, maximum height? a ball is thrown upward with an preliminary speed of 29.4 m/s. What maximum height does the ball attain? while an merchandise is thrown upward, it decelerates 9.8 m/s each and each 2nd, until its speed = 0 m/s. right now the article is at its maximum element. Then it hastens downward at 9.8 m/s each and each 2nd. If the beginning up element and end element are the comparable, the path is symmetrical. the article strikes down the comparable distance it moved up. the mind-blowing speed on the top of the downward holiday equals the preliminary speed of the upward holiday. The time to realize the circulate as much as the utmost element is the comparable because of the fact the time to fall backtrack to the beginning up element. the linked fee of a ball with preliminary vertical speed of 29.4 m/s will shrink 9.8 m/s each and each 2nd until the vertical speed = 0 m/s vf = vi + a*t vf = 0 vi = 29.4 a = -9.8 0 = 29.4 +(-9.8 * t) t = 3 seconds because of the fact the linked fee is lowering at a persevering with fee, the area the ball strikes equals its conventional speed circumstances the time. conventional speed = (preliminary speed + very final speed) ÷ 2 conventional speed = (+29.4 + 0) ÷ 2 = 14.7 m/s Time = 3 seconds Distance = 14.7 m/s * 3 s = 40 4.a million m is it 44m or 88 m (40 4) are you able to apply d=vt NO, because of the fact the linked fee replaced into no longer 29.4 m/s each and all the time!! v commonly skill consistent speed. while an merchandise is shifting upward, its speed is lowering because of the attractive rigidity between the article and the earth. are you able to locate t? sure, by way of dividing the preliminary speed by way of the deceleration

Answer 3

We know that:
a = -9.8

int(a dt) = int(-9.8 dt)
v = -9.8t + C
v(0) = 10 = -9.8(0) + C
C = 10
v = -9.8t + 10

Now we must find the time when the ball stops, when the velocity is 0, as this is obviously at the maximum height.

v = 0 = -9.8t + 10
t = 10 / 9.8

int(v dt) = int(-9.8t dt + 10 dt)
h = (-9.8 / 2)t^2 + 10t + C

Assuming that the initial height is 0.
h(0) = 0 = (-9.8 / 2)0^2 + 10(0) + C
C = 0

h = (-9.8 / 2)t^2 + 10t
h(10 / 9.8) = (-9.8 / 2) (10 / 9.8)^2 + 10(10 / 9.8)
h(10 / 9.8) = -(100 / (9.8 * 2)) + (100 / 9.8)
h(10 / 9.8) = 100 / 9.8
h(10 / 9.8) ~= 10.204

Therefore, the maximum height reached by the object is 10.204 metres. Note that the mass was a red herring as acceleration due to gravity is not a function of mass.

Answer 4

mass has no effect

Vf^2 = 2ad + Vi^2

0^2 = 2(-9.8)d + 10^2
-100 = -19.6d
d = 5.1m

Answer 5

v^2 - u^2 = 2.g.h or h = u^2/2g

h = 100/19.6 = 5.1 m

The ball will reach 5.1 m heigh before it starts falling back to ground.

Answer 6

about 24 meters

Answer 7

5.10 meters