Question

PLS ANSWER FAST.;........ URGENT

A ball is thrown vertically up from a point A on the ground with a velocity 11m/s. The ball rises up to the top most point B and then comes back down to a point C. All this happens while the ball moves only under the influence of gravity with a uniform acceleration of 10 m/s2 downwards. When it reaches C, the ball has a velocity 9 m/s pointing downwards. How high above A is C?

Answer 1

So a ball is thrown from a point A upwards with a velocity 11 m/s and is reaches maximum height. It further falls downwards and reaches a point C where its velocity is 9 m/s. Acceleration due to gravity is given as 10 m/s².

Using the equation, taking A as initial point and B as final

$\large\black\boxed{v^2-u^2=2as}$

u = 11 m/s

v = 0 (since at maximum height, velocity is zero)

a = -10 m/s² (acceleration due to gravity acts against the motion of ball)

s₁ = ? (distance between A and B)

$0^2-11^2 = 2*(-10)*s_1$

$s_1 = 6.05m$

That is, B is 6.05 m above the ground.

Now, using same equation and taking B as initial point and C as final point.

u = 0 (at B)

v = 9 m/s (at C)

a = 10 m/s² (acceleration due to gravity acting in the direction of motion of ball)

s₂ = ? (distance between B and C)

$9^2-0^2 = 2*10*s_2$

$s_2 = 4.05m$

So we need to find distance between A and C. That is actually the difference between s₁ and s₂

s₁ - s₂ = 6.05 - 4.05 = 2 m.

So, C is 2 m above A.