Question

NEED HELP DUE AT 11:59!! A ball is thrown horizontally from the top of
a building 130 m high. The ball strikes the
ground 53 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.

Answer 1

Answer:

Since the ball was thrown horizontally, there was no vertical component in that force. and hence, the initial vertical velocity of the ball is 0 m/s and the initial horizontal velocity is r.

We are given:

initial velocity  (u) = 0 m/s     [vertical]

final velocity (v) = v m/s  [vertical]

time taken to reach the ground (t) = t seconds

acceleration (a) = 10 m/s/s   [vertical , due to gravity]

height from the ground (h) = 130 m

displacement (s) = 53 m [horizontal]

Solving for time taken:

From the third equation of motion:

s = ut + 1/2 at²

130 = (0)(t) + 1/2 * (10) * t²

130 = 5t²

t² = 26

t = √26 seconds  or   5.1 seconds

Final Horizontal velocity of the ball

Since the horizontal velocity of the ball will remain constant:

the ball covered 53 m in 5.1 seconds [horizontally]

horizontal velocity of the ball = horizontal distance covered / time taken

Velocity of the ball = 53 / 5.1

Velocity of the ball = 10.4 m/s

Answer 2

Answer:

51.51519 m/s

Explanation:

Given: $a_{x} =0$ $a_{y} -g$ $v_{yo} =0$ $x_{o} =0$ $x=53$$y_{o} =130$

X-direction                           | Y-direction

$x=x_{o} +v_{xo}t$                         | $y=y_{o} +v_{yo}t+\frac{1}{2}a_{y}t^2$

$53=0v_{xo}(5.15078)$                 | $0=130+\frac{1}{2}(-9.8)t^2$

$53=v_{xo} (5.15078)$                    | $-130=-4.9t^2$

$\frac{53}{5.15078} =v_{xo}$                             |  $\sqrt{\frac{-130}{-4.9} }=\sqrt{t^2}$

$10.2897=v_{xo}$                            | $5.15078=t$

$v=\sqrt{v_{y}^2+ v_{x}^2}$                            | $v_{y}^2 =v_{yo}+2a_{y} d$

$v=\sqrt{(50.27771)^2+(10.2897)^2}$ | $\sqrt{v_{y}^2} =\sqrt{2(-9.8)(0-130)}$

$v=51.51519 m/s$                        | $v_{y}=50.47771$