a certain ball is thrown vertically into the air from ground level. At the instant that the ball is traveling

Question

a certain ball is thrown vertically into the air from ground level. At the instant that the ball is traveling upward at 15 m/sec, the ball is 10 meters above the ground. The initial speed of the ball was most nearly ( ? ) m/sec? please help!!!!

Answer 1

Consider an equation of motion
S=.5gt^2
We know that the time it takes for the Ball to travel 10m
t=sqrt(2S/g)

V=Vo - gt
V0=V+gt
V0=V+g sqrt(2S/g) or
V0=V+ sqrt(2gS/)
V0=15 + sqrt(2 x 9.81 x 10 )
V0=29.0 m/s

Answer 2

ignoring the height of the ball at release (probably about 2 meters)

Vf^2 = Vi^2 + 2 a x

(15 m/s)^2 - 2 x (-9.8 m/s^2) x 10 m = Vi^2

225 m^2/s^2 + 196 m^2/s^2 = Vi^2 = 421 m^2/s^2

so Vi = 20.5 m/s

of course if you wanted to include the 2 meters then 10 m would really be 8 m and the answer would be 19.5 m/s

so say the answer is about 20 meters / second

UPDATE:

dm_cork

you forgot the 2 whilst multiplying
v^2 = u^2 + (2) x as


Edward.

S = Vot + 1/2 a t ^2.
you forgot the Vot term.

Answer 3

a) s = ut + 0.5at^2, anticipate no air resistance so a = 0. s = ut so u = s / t = 26 / 3 = 8.7m/s b) s = ut + 0.5at^2, this time a is acceleration by using gravity, lower back anticipate no air resistance. Rearrange for u. u = s / t - 0.5at = 10 / 3 - 0.5 * -9.8 * 3 = 3.3 + 14.7 = 18.0 m/s c) preliminary speed is basic pythagoras, speed^2 = x speed^2 = y speed^2 speed^2 = 8.7^2 + 18.0^2 = seventy 5.a million + 325.2 = 4 hundred.3 speed = 20.0 m/s d) lower back pythagoras. tan theta = y speed / x speed = 18.0 / 8.7 = 2.a million theta = sixty 4.3 stages e) on the dazzling of the balls direction the y speed = 0 v^2 = u^2 + 2as whilst v = 0 we are able to write 2as = -(u^2) and s = -(u^2 / 2a) s = -(18.0^2 / (2 * -9.8)) = sixteen.6m I even have rounded all written numbers to a million decimal place, yet used the whole numbers in calculations.

Answer 4

The motion is under constant acceleration, that due to gravity. Taking upwards as being positive with respect to distance, the acceleration is downward, i.e. negative. The equation you need is

v^2 = u^2 +2as

where v is the speed at the end point, u is the speed at the beginning, s is the distance travelled between those two times and a is the accelerations.

So v = 15 m/sec
u = ? (use u in the formula)
a = -9.8 m/sec^2 (gravity)
s = 10 m

This gives u = 17.8 m/sec.

Answer 5

At the moment that I am responding to you, there are two other responses with different numerical answers.
You have v1=?, d=10 m, v2= 15 m/s, a = -9.8 m/s

v2^2 - v1^2 = 2ad

Therefore v1^2=v2^2-2ad. Substitute and see which is correct!

(Other people use s for distance, and g (in this case) for the acceleration.