Question

A ball is thrown from a top of a building of height 20m. If the initial velocity of the ball is 15m/s at 370 above the horizontal find, a) the velocity of the ball at 2s. b) the position at which the ball strike the ground.​

Answer 1

Explanation:

Givens:

$h = 20 \: m$

$v _{ix} = 15 \cos(37) = 11.97$

$v _{iy} = 15 \sin(37) = 9.03$

$a _{y} = - 9.8$

Let t=2 be the final velocity. Since Velocity stays the same horizontal , acceleration due to gravity changes the vertical direction

Use this equation

$v _{y} = v _{iy} + a _{y}(t)$

Plug in the knowns

$v _{y} = 15 \sin(37) +2 ( - 9.8)$

$v _{y} = - 10.57$

So the velocity in the y direction at 2 seconds is -10.57

The velocity in the x direction at 2 seconds is 11.97

Use the Pythagorean theorem to find the total velocity

$v = \sqrt{( - 10.57) {}^{2} + (11.97) {}^{2} }$

$v = 15.97$

b. The range at which it strike the ground.

We need to find when the velocity at the top is zero.

Using the y direction,

$0 = v \sin( 37 ) - 9.8(t)$

$t = 0.921$

Next, find the height of the max.

$h = \frac{1}{2} ( 9.8)(0.921) {}^{2} = 4.16$

So the total distance is 4.16+20= 24.16

Next to find the total time it falls

$24.16 = \frac{1}{2} (9.8) {t}^{2}$

$t = 2.2$

So our total flight time is

$3.141$

Range is

$v \cos(37) (3.141) = 37.63$