A man can throw a ball a maximum horizontal distance of 73.4 meters. the acceleration is 9.8 m/s^2. How far can he throw the same ball vertically upward at the same initial speed?
d = Vo^2*sin(2A)/g = 73.4 m.
A = 45o @ max hor. distance.
Vo^2*sin(90)/9.8 = 73.4.
Vo^2/9.8 = 73.4
Vo = 26.8m/s[45o].
Yo = 26.8*sin45 = 19m/s. = Ver. component.
Y^2 = Yo^2 + 2g*h.
0 = 19^2 - 19.6h. h = ?.
To determine how far the man can throw the ball vertically upward, we need to make use of the same initial speed and acceleration.
First, we need to understand that the horizontal and vertical motions of the ball are independent of each other. This means that the distance the ball travels horizontally has no impact on the distance it travels vertically.
Given that the maximum horizontal distance the ball can travel is 73.4 meters, this is solely determined by the initial speed and acceleration. Since there is no acceleration in the horizontal direction, we can assume the initial speed is the same as the maximum horizontal speed.
To find the initial speed, we need to use the fact that acceleration due to gravity is 9.8 m/s^2. We can use the following equation to calculate the initial speed (vi) when the projectile reaches its maximum height:
v² = u² + 2as
Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (the speed we want to find)
a = acceleration due to gravity (-9.8 m/s^2)
s = displacement (the height reached, which we don't yet know)
Rearranging the equation, we get:
0 = u² + 2as
Now, we can substitute the known values into the equation:
0 = u² + 2(-9.8 m/s^2)(s)
Since the ball reaches its maximum height when it comes to rest (0 m/s), we can eliminate the final velocity term.
0 = u² - 19.6s
Next, we need to find the value of s. In this case, s represents the maximum vertical height reached by the ball, which is half the total vertical distance traveled (since the ball will come back down to the same level).
Therefore, s = 73.4 m / 2 = 36.7 m
Substituting this value into our equation, we have:
0 = u² - 19.6(36.7 m)
Now, we can solve for the initial vertical speed (u):
u² = 19.6(36.7 m)
u² = 716.12 m²/s²
Taking the square root of both sides:
u ≈ 26.78 m/s
Therefore, the man can throw the same ball vertically upward at approximately 26.78 m/s.