Question

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?

Answer 1

Answer:  well the time it takes to fall 100m is the same time it takes to travel 65m horizontally.  

The time to fall vertically, t is sqrt(2d/g). [this comes from d = 1/2at^2]  

so t = sqrt(2*100/9.8) = 4.52s.  

The vertical speed at that time is g*t = 9.8*4.52 = 44.3m/s  

The horizontal speed is the horizontal distance over the same 4.52s, = 65/4.52 = 14.4m/s.  

so the final velocity is = sqrt(44.3^2 + 14.4^2) = 46.6m/s

Explanation: yes

Answer 2

Answer:

v = 46.55 m/s

Explanation:

It is given that,

A ball is thrown horizontally from the top of a building 0.10 km high, d = 0.1 km = 100 m

The ball strikes the ground at a point 65 m horizontally away from and below the point of release, h = 65 m

At maximum height, velocity of the ball is 0. So, using the equation of motion as :

$d=ut+\dfrac{1}{2}at^2$

Here, a = g

$100=0+\dfrac{1}{2}\times 9.8t^2$

$t=4.51\ s$

Let $v_x$ is the horizontal velocity of the ball. It is calculated as :

$v_x=\dfrac{65\ m}{4.51\ s}=14.41\ m/s$

Let $v_y$ is the final speed of the ball in y direction. It can be calculated as :

$v_y^2+u_y^2=2as$

$u_y=0$

$v_y^2=2gd$

$v_y^2=2\times 9.8\times 100$

$v_y=44.27\ m/s$

Let v is the speed of the ball just before it strikes the ground. It is given by :

$v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{14.41^2+44.27^2}$

v = 46.55 m/s

So, the speed of the ball just before it strikes the ground is 46.55 m/s. Hence, this is the required solution.