# Find the time it takes the stone to catch up with the ball and find the velocities of the stone and the ball when they are at the same height.A ball is thrown vertically upward with an initial...

Find the time it takes the stone to catch up with the ball and find the velocities of the stone and the ball when they are at the same height.

A ball is thrown vertically upward with an initial speed of 11m/s. One second later, a stone is thrown vertically upward with an initial speed of 25m/s.

sociality | High School Teacher | (Level 1) Valedictorian

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We have to find the time it takes the ball and the stone to meet and their velocities when they meet.

Now the distance traveled in time t, by an object with an initial velocity u and which has an acceleration a, is given by the expression (2u + at)* t/2.

Let's take the acceleration due to gravity as 9.8 m/s^2 acting downwards. The ball is thrown up at 11m/s and a second later the stone is thrown up at 25m/s.

When the two meet, they have traveled to an equal height. So we get the equation:

(2*11 - 9.8*t)*t/2 = (2*25 – 9.8*(t-1))*(t-1)/2

=> 11t – 4.9t^2 = 25(t-1) – 4.9(t-1) ^2

=> 11t – 4.9t^2 = 25t – 25 – 4.9t^2 – 4.9 + 9.8t

=> 11t – 4.9t^2- (25t – 25 – 4.9t^2 – 4.9 + 9.8t) = 0

=> 11t – 4.9t^2 – 25t + 25 + 4.9t^2 + 4.9 – 9.8t =0

=> 11t– 25t + 25 + 4.9 – 9.8t =0

=> t = (-25 - 4.9) / (11 - 25 - 9.8)

=> t = 1.256 s

Therefore they reach an equal height 1.256s after the ball is thrown.

The velocity of the ball when they meet is 11- 9.8*1.256= 1.308m/s acting downwards and that of the stone is 25 - 9.8*.256 = 22.49 m/s acting upwards.

Therefore they meet after approximately 1.256 s of the ball being thrown.

When they meet the velocity of the ball and the stone are 1.308m/s acting downwards and 22.49 m/s acting upwards, resp.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The distance travelled by an an object thrown upwards is given by the formula:

s = ut - (gt^2)/2

Where:

s = Distance travelled

u = Initial upward speed

t = Time elapsed

g = Acceleration due to gravity = 9.8 m/s^2

Using this formula the distance travelled by the ball at time t is given by:

s(ball) = 11t - (gt^2)/2

As the stone is thrown one second after the ball, the time applicable to the stone, as compared to that applicable to ball will be (t - 1).

Therefor the distance travelled by the stone at time t is given by:

s(stone) = 25(t -1) - [g(t - 1)^2]/2

= 25t - 25 - g(t^2 - 2t + 1)/2

= 25t - 25 - (gt^2 - 2gt + g)/2

= 25t - 25 - (gt^2)/2 + gt - g/2

When the stone and ball are at the same height:

s(ball) = s(stone)

Substituting the above expressions of distance for ball and stone in above equation:

11t - (gt^2)/2 = 25t - 25 - (gt^2)/2 + gt - g/2

==> 25t - 11t - 25 - (gt^2)/2 + (gt^2)/2 + gt - g/2

==> 14t - 25 + gt - g/2 = 0

==> 14t + gt = 25 + g/2

==> t(14 - g) = 25 - g/2

==> t = (25 + g/2)/(14 + g)

Substituting value of g in above equation:

==> t = (25 + 9.8/2)/ (14 + 9.8) = 29.9/23.8 = 1.2563 s

Time taken by stone to catch up with ball

= t - 1 = 1.2563 - 1 = 0.2563 s

Velocity of ball at 1.2563 seconds = 11 - 9.8*1.2563 = - 1.3118 m/s

Velovity of stone at 1.2563 seconds = 25 - 9.8*0.2563 = 22.4880 m/s