Answer:
Just before it strikes the ground, the speed of the ball is 46.4 m/s.
Step-by-step explanation:
Hi there!
The equation of position and velocity of the ball at a time t are the following:
r = (x0 + v0 · t, h0 + 1/2 · g · t²)
v = (v0, g · t)
Where:
r = position vector of the ball at a time t.
x0 = initial horizontal position of the ball.
v0 = initial velocity.
t = time.
h0 = initial height.
g = acceleration due to gravity.
v = velocity of the ball at a time t.
Let´s place the origin of the frame of reference on the ground at the edge of the building. The initial position of the ball is r0 = (0, 100) m.
Then, the final position of the ball will be r1 = (65, 0) m.
Using the equation of the vertical component of the position vector, we can find the time at which the ball hits the ground (height = 0 m):
h = h0 + 1/2 · g · t²
0 m = 100 m - 1/2 · 9.8 m/s² · t²
-100 m / -4.9 m/s² = t²
t = 4.5 s
We know that in 4.5 s the ball travels 65 m in the horizontal direction. Then, using the equation of the horizontal component of the position vectro, we can find the initial horizontal velocity, v0:
x = x0 + v0 · t (x0 = 0)
x = v0 · t
x/t = v0
65 m / 4.5 s = v0
v0 = 14.4 m/s
Now, let´s calculate the vertical component of the velocity vector at t = 4.5 s:
vy = g · t
vy = -9.8 m/s² · 4.5s
vy = -44.1 m/s
When the ball hits the ground, its velocity vector will be:
v = (14.4, -44.1) m/s
The magnitude of this vector is calculated as follows:
|v| = √[(14.4 m/s)² + (-44.1 m/s)²] = 46.4 m/s
Just before it strikes the ground, the speed of the ball is 46.4 m/s.