vector A has magnitude 12 m and direction +y
so we can say
vector B has magnitude 33 m and direction - x
Now the resultant of vector A and B is given as
now for direction of the two vectors resultant will be given as
so it is inclined at 160 degree counterclockwise from + x axis
magnitude of A and B will be
so magnitude will be 35.11 m
Answer:
His pitching speed is 38 m/s.
Explanation:
Hi there!
Please see the attached figure for a better understanding of the problem.
The position of the ball at any time t is given by the following vector:
r = (x0 + v0 · t, y0 + 1/2 · g · t²)
Where:
r = position vector of the ball at time t.
x0 = initial horizontal position.
v0 = initial horizontal velocity.
t = time.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Let's place the origin of the frame of reference at the throwing point so that x0 and y0 = 0.
When the ball reaches the ground, its position vector will be r1 (see figure). Using the equation of the vertical component of the position vector, we can find the time at which the ball reaches the ground. At that time, the horizontal component of the position is 30 m and the vertical component is -3.0 m (see figure):
y = y0 + 1/2 · g · t² (y0 = 0)
y = 1/2 · g · t²
-3.0 m = 1/2 · (-9.8 m/s²) · t²
-3.0 m / -4.9 m/s² = t²
t = 0.78 s
Now, knowing that at this time x = 30 m, we can find v0:
x = x0 + v0 · t (x0 = 0)
x = v0 · t
30 m = v0 · 0.78 s
v0 = 30 m / 0.78 s
v0 = 38 m/s
His pitching speed is 38 m/s.
