A man can throw a ball a maximum horizontal
distance of 149 m.
The acceleration of gravity is 9.8 m/s2 .
How far can he throw the same ball verti-
cally upward with the same initial speed?
Answer in units of m.
If you think about it, the total distance travled in the x direction is equal to the y direction since to get the max an object can go in the x direction is acquired at a 45 degree angle. So you are going a total of 149 m, at the highest point the distance traveled will be 149/2, because distance up always equals distance down. So the answer is 149/2 or 74.5 m.
To find the maximum vertical distance the man can throw the ball upward, we can use the concept of projectile motion.
In projectile motion, the vertical and horizontal motions are independent of each other. The time taken for the ball to reach its maximum height when thrown vertically upwards is the same as the time taken for the ball to fall back down to the ground when thrown horizontally.
Using the kinematic equation for vertical motion, we can find the time taken for the ball to reach its maximum height (the time it takes for the ball to reach its highest point):
v = u + at
Here,
v = final velocity (0 m/s at the highest point)
u = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken
Since the ball reaches its maximum height at the highest point, its final velocity is 0 m/s. Thus, we can rewrite the equation as:
0 = u - 9.8t
Solving for t, we get:
t = u / 9.8
The total time of flight for the ball (from going up to coming back down) is twice this time, so:
Total time of flight = 2 * (u / 9.8)
Now, using the horizontal distance (149 m) the man can throw the ball, we can find the total time of flight:
Horizontal distance = horizontal velocity * total time of flight
Since there is no horizontal acceleration (horizontal velocity remains constant), the horizontal distance is equal to the horizontal velocity (unknown) multiplied by the total time of flight.
149 m = horizontal velocity * (2 * (u / 9.8))
Simplifying, we have:
149 m = 2u / 9.8
Solving for u, we get:
u = 149 m * 9.8 / 2
u ≈ 725.1 m/s
Now that we know the initial vertical velocity (u), we can find the maximum vertical distance reached by the ball using the time taken to reach the highest point:
Vertical distance = (u^2) / (2 * g)
Substituting the values, we have:
Vertical distance = (725.1 m/s)^2 / (2 * 9.8 m/s^2)
Vertical distance ≈ 26,456.5 m
Therefore, the man can throw the same ball vertically upward for a maximum distance of approximately 26,456.5 meters.
To find the maximum vertical distance the same ball can be thrown upward, we need to understand the projectile motion and the relationship between horizontal and vertical components.
In projectile motion, an object moves under the influence of gravity with both horizontal and vertical components. The horizontal motion is unaffected by gravity and remains constant, while the vertical motion is influenced by gravity and follows a parabolic path.
Given that the maximum horizontal distance is 149 m, we can assume that the same initial velocity is applied for both the horizontal and vertical components.
In projectile motion, the vertical distance reached by an object thrown upwards can be determined using the equation:
h = (v^2 * sin^2(θ)) / (2g)
Where:
h = vertical distance
v = initial velocity of the object
θ = launch angle (angle between the initial velocity and the horizontal)
g = acceleration due to gravity (9.8 m/s^2)
Since the initial velocity and launch angle are the same for both horizontal and vertical throws, we can use the same values for v and θ.
To find the initial velocity (v), we need to consider the horizontal distance:
v = d / t
Where:
d = horizontal distance (149 m)
t = time taken
The time taken for the ball to reach the maximum horizontal distance is the same as the time taken for the ball to reach its maximum height. Hence, the vertical component of the motion will take the same time as the horizontal component.
Therefore, we find the time taken by using the equation:
t = d / v
By substituting the given values, we can find v:
v = 149 m / t
Now, let's calculate t:
t = d / v
t = 149 m / 149 m/s
Simplifying, we find t = 1 second.
Now, we can substitute the values into the equation for h:
h = (v^2 * sin^2(θ)) / (2g)
h = ((149 m/s)^2 * sin^2(θ)) / (2 * 9.8 m/s^2)
h = (22101 m^2/s^2 * sin^2(θ)) / 19.6 m/s^2
Now the answer depends on the launch angle (θ). If we assume the ball is thrown straight up (θ = 90 degrees), we get:
h = (22101 m^2/s^2 * sin^2(90)) / 19.6 m/s^2
h = (22101 m^2/s^2 * 1) / 19.6 m/s^2
h ≈ 1127.03 m
Therefore, if the ball is thrown straight up with the same initial speed, it can reach a maximum vertical distance of approximately 1127.03 meters.