Question

A ball is dropped from a height. If it takes 0.200 s to cross the last 6m before hitting the ground, find the height from which it was dropped.

Please answer this fast. No spams please.

Answer 1

Answer:

Answer is 48.05 m

Explanation:

Lets take velocity as u at point P

then,

$s=ut+1/2at^{2} \\6=u(0.2)+1/2(10)(0.2)^2\\6=0.2u+0.2\\u=29 ms^-1\\\\\\</p><p>Now using v=\sqrt{2gh}$

$29^{2} = 2g(H-6)\\42.05 = H-6\\H=48.05 m$

Answer 2

Answer:

Answer:

Answer is 48.05 m

Explanation:

Lets take velocity as u at point P

then,

s=ut+1/2at^{2}

6=u(0.2)+1/2(10)(0.2)^2

6=0.2u+0.2

u=29ms

Now using v={2gh}.

29^{2} = 2g(H-6)

42.05 = H-6

H=48.05