a girl throws a ball vertically downward at 10 m/s from the roof of the building 50m high.how long will it...?

Question

how long will it take the ball to reach the ground?what will be its velocity when it strikes the ground?

Answer 1

This is a one dimensional projectile motion problem which will employ two formulas to solve...

d = v_i * t + 1/2 at^2
v = a*t + v_i
where d is the distance the ball travels after time, t, experiencing a constant acceleration, a, with an initial velocity v_i.
Velocity = acceleration * time

We know the building is 50 m tall, so d = 50.
We know the ball has an initial velocity of 10 m/s, so v = 10 m/s.
We know gravity will cause the ball to accelerate at a constant rate of 9.81 m/s^2, so a = 9.81 m/s^2.
Plug into the formula and solve for t (using the quadratic equation if necessary).

After we know the time, and since we already know the constant acceleration the ball experiences, finding the final velocity is no problem,
v = a*t + v_i
plug in and solve.

For the time, t, I computer the answer to be: t = 2.33 seconds.
For the final velocity, v, I found the answer to be: v = 32.86 m/s.

Answer 2

It will be 5 seconds when it hits the ground because if the ball goes 10 miles every second then 50 divided by 10 is 5 so it would be 5 and the velocity would be 2 points

Answer 3

ok so the initial speed is 10m/s down. and it accelerates at a constant rate of 9.8m/s^2. so because the rate of acceleration is constant, the average speed would just be the initial + final /2.

and if you multiply the time by average speed, u get the distance travelled.
so (10m/s + Vf)/2 * t = 50m
(10m/s + Vf) * t = 100m

now, you also know that the acceleration would be 9.8, which is equal to the final speed - initial speed / time
9.8 = Vf-10/t (change in speed over time)

so u have 2 equations and 2 unknowns, it isnt hard to solve it.
9.8t = Vf-10-------> Vf = 9.8t + 10
10t + Vf * t = 100m
10t + (9.8t + 10)t = 100
10t + 9.8t^2 + 10t = 100
9.8t^2 + 20t - 100 = 0

quadratic equation. use the quadratic equation to solve for the time.

t = [-b +- sqrt (b^2 - 4ac)]/2a
= [-20 +- sqrt (20^2 - 4(9.8)(-100)] / 2(9.8)
= [-20 +- sqrt (400 + 3920)] / 19.6
= 2.333s is how long it takes to hit the ground

so then Vf = 9.8t + 10
= 32.86335 m/s is the final speed when it hits the ground

and there u go. hope that helps

Answer 4

Initial speed "u"is 10m/s.

Acceeration "a" is 9.8 m/s^2.

Distance traveled is 50m.

Time, t =?

Final velocity,v =?

v^2 = u^2 + 2 as = 10 x10 + 2 x 9.8 X 50. = 1080

v = 32.86(3353) m/s.

Distance = Average speed x time.

Therefore, time "t" = distance / Average speed .

= 50 / {(32.86 +10) / 2 }

= 2.33(29953) s.

Answer 5

on earth the gravitational acceleration rate: 9.8067meters per second per second or 9.8067m/s squared,each second u would have to increase by that and if were at sea level.balls mass and wind resistance and was 10m/s constant or thrown as 10 m/s was final velocity

Answer 6

it will take 5 sec to reach the ground.

Answer 7

6 sec

Answer 8

5 sec. and 2 points for me

Answer 9

It will take 2.32 seconds and its velocity will be 33.2 metres/second.