Question

1 pts Question 6 6) A ball is launched straight upwards at 12 m/s from the side of a building which is 101 meters above the ground. At the exact same time, a projectile is launched from the ground (the base of the building). The projectile is 55 meters from the base of the building. If the projectile strikes the other ball 20 meters above the ground, at what angle, in degrees, does it hit the other ball? If below horizontal, include a negative sign. Vil Vi2

Answer 1

101m 20m

Consider the motion of the first ball. we will calculate the time in which the height of the ball is 20m above the ground. The initial height of the first ball is

ho = 101m

The height of the ball when hit by the second ball is

h = 20m

The initial velocity of the ball is

Uli = 12 m/s

The acceleration of the first ball is equal to the gravitational acceleration of the Earth

a = -9.8 m/s

The acceleration is uniform, the motion is governed by the kinematics equations for uniform motion. We use

h = h, + Ulit + at

where t is the time in which the ball reaches 20m height.

20 = 101 + 12+ + + X(-9.8x)t?

4.97 - 12t -81 = 0

This is quadratc equation in time. We know that the roots of quadratic equation are

ar? + bc+c=0

-63 - 4ac 2a

We get

$t= \dfrac{-(-12) \pm \sqrt{(-12)^2-4\times 4.9\times (-81)}}{2\times 4.9}$

t = -3.022s, 5.4715

We are interested in time after the ball was thrown, we are interested in positve time. Therefore,

Consider the motion of the second ball. The acceleration in the horizontal direction is zero. The distance covered along the horizontal direction is given by

$x = speed \times time$

where speed is the horizontal speed of the ball and x should be 55m in order for the second ball to hit the first ball.

55m = (viz cose) x 5.4715

$v_{i2}\cos\theta =\dfrac{55m}{5.471s}$

$\Rightarrow v_{i2}\cos\theta =10.05\ m/s\ \ \ \ \ \ \ \ \ \ (1)$

The acceleration of the second ball in the vertical direction is equal to the gravitational acceleration of the Earth. Applying kinematics equation for the vertical direction.

$h = v_{initial}t + \dfrac{1}{2} at^2$

n = (Ui2 sin )t + -at-

We want that at time t=5.471s, the height of the second ball must be 20m.

$20 = (v_{i2}\sin\theta)5.471 + \dfrac{1}{2}\times (-9.8)\times 5.471^2$

$(v_{i2}\sin\theta)5.471 = 20-\dfrac{1}{2}\times (-9.8)\times 5.471^2$

$\Rightarrow v_{i2}\sin\theta= 30.46\ m/s\ \ \ \ \ \ \ \ \ \ (2)$

We want to find the vertical component of the second ball's velocity at the moment before it hit the first ball. If $v_{f2}$ is the magnitude of the second ball's velocity when it hit the first ball, the vertical component of the second ball's velocity is $v_{f2}\sin\phi$. The vertical motion is uniformly acclerated, we use

$v_{f2}\sin\phi = v_{i2}\sin\theta - 9.8 \times 5.471$

Using equation (2)

$v_{f2}\sin\phi = 30.46 - 9.8 \times 5.471$

$v_{f2}\sin\phi = -23.15\ m/s\ \ \ \ \ \ \ \ \ \ (3)$

The horizontal component of the final velocity of the second ball is equal to the horizontal component of its initial velocity because the acceleration in the horizontal direction is zero.

$v_{f2}\cos\phi = v_{i2}\cos\theta$

Using equation (1)

$v_{f2}\cos\phi = 10.05\ m/s\ \ \ \ \ \ \ \ \ \ (4)$

Dividing equation (4) by (3) we get

$\dfrac{v_{f2}\sin\phi}{v_{f2}\cos\phi} = \dfrac{-23.15}{10.05}$

$\tan\phi = \dfrac{-23.15}{10.05}$

$\phi = \tan^{-1}\left (\dfrac{-23.15}{10.05} \right )$

we get

$\boxed{\phi = -66.5^o}$