A ball is thrown verticall upward with a velocity of +10.0m/s How high does it rise?

Question

A ball is thrown verticall upward with a velocity of +10.0m/s How high does it rise?

Hi, I asked this question yesterday and I got the answer, its 5m, but im not entirely sure how they got it. Somehow they solved for t, and they used like -10m/s^2 for the acceleration I dont understand where they got the acceleration, as its not presented in the problem, and I dont think I have an equation that gives me acceleration from velocity. It seems like a really simple question but im stumped, any help? thanks

Answer 1

u = initial velocity
v = final velocity
a = acceleration = -gravitational force = -9.81ms^-2 or -10ms^-2 (taking upwards as positive)

using formula...

v^2 = u^2 + 2ad

to reach max height, v = 0,

0 = 10^2 + 2(-10)d
20d = 100
d = 5m

Answer 2

Use the formula

v^2 = u^2 + 2as

Where v is the final velocity (zero in this case when the ball comes to rest at the top of its path)

u is the initial velocity (+10m/s in this case)
a is the acceleration due to gravity (-9.8m/s)
s is the displacement (the maximum height in this case and what you want to find)

0=10^2+2x(-9.8)s

0 = 100-19.6s
19.6s=100
s=100/19.6
s=5.1 metres

Answer 3

properly first of all you need to do not forget that there are 4 linear % equations , that incorporates 4 uncomplicated characteristics : initial speed ( u ) , most suitable speed (v) , initial acceleration (u), time of action (t) and entire distance coated (s) .The 4 equation contain 4 unknowns, you are able to wish to attraction to close 3 in the previous you'll discover the fourth. between the most equation is v^2 = u^2 + 2as because on the right of the trajectory of the ball, the speed will in all probability be 0, we account properly worth of v to be 0. 0 = u^2 + 2as Making s area of the equipment, it to that end commerce to : (0- (u^2) )/ (2a) = s ([0- (80ft/s ^2)) ] / (2 * (-32 ft/sec^2) =s s = ( -6400 ft/s) / -sixty 4 (ft/(s^2) ) s= (-6400/-sixty 4) m s=100 ft the cost is a useful one, which make sure that the answer is ideal , as distance are not any further waiting to be poor the following.

Answer 4

a = dv/dt = -10m/sec^2 This is acceleration due to gravity.
v= -10t+C
when t = 0, v = 10m/s, so C= 10
v=-10t+10
v=ds/dt
s = -5t^2+10t +C
when t = 0 s = 0 so,
s=-5t^2 +10t = height of ball
The ball is at max height when t=-10/2(-5) = 1
s max = -5(1)^1 + 10 = -5+10 = 5

Answer 5

(1/2)mv^2 = mgh, by the law of energy conservation.

Solve for h,
h = v^2/(2g) = 10^2/(2*9.8) = 5.1 m