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:

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.

Answer 2

The speed of the ball just before it strikes the ground is equal to 46.55 m/s.

Given the following data:

• Horizontal distance = 65 meters

• Height of building = 0.10 km = 100 meters

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 $m/s^2$.

To determine the speed of the ball just before it strikes the ground:

First of all, we would determine the time it took the ball to strike the ground by using the formula for maximum height.

$H = \frac{1}{2} gt^2\\\\100 = \frac{1}{2} \times 9.8 \times t^2\\\\200 = 9.8t^2\\\\t^2 = \frac{200}{9.8} \\\\t^2=20.41\\\\t=\sqrt{20.41}$

Time, t = 4.52 seconds

Next, we would find the horizontal velocity:

$Horizontal\;velocity = \frac{horizontal\;distance}{time} \\\\Horizontal\;velocity = \frac{65}{4.52}$

Horizontal velocity, V1 = 14.38 m/s

Also, we would find the velocity of the ball in the horizontal direction:

$V_2^2 = U^2 + 2aS\\\\V_2^2 = 0^2 + 2(9.8)(100)\\\\V_2^2 = 1960\\\\V_2 = \sqrt{1960} \\\\V_2 = 44.27 \;m/s$

Now, we would calculate the speed of the ball just before it strikes the ground by finding the resultant speed:

$V = \sqrt{V_1^2 + V_2^2} \\\\V = \sqrt{14.38^2 + 44.27^2}\\\\V = \sqrt{206.7844‬ + 1959.8329‬}\\\\V =\sqrt{2166.6173}$

Speed, V = 46.55 m/s

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