Question

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What far vertically will it travel before hiting the ground A. 40 m B. 30 m C. 60 m D. 50 m

Answer 1

The ball will travel approximately 50 m vertically before hitting the ground, considering the time to reach the peak height and the symmetrical trajectory of its motion.

Step-by-step explanation:

The ball launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s subject to gravitational deceleration will have a symmetrical trajectory. Since the vertical speed is 30 m/s, we can calculate the time it takes for the ball to reach its peak height, where its vertical velocity will become zero due to the acceleration due to gravity (9.8 m/s2). The time to reach maximum height (t) can be found using the equation vertical speed = gravity × time, which gives us t = vertical speed/gravity. Plugging in the numbers, we get t = 30 m/s / 9.8 m/s2 which is approximately 3.06 seconds to reach the peak. Since the total flight time will be double this (as it takes the same amount of time to come down as it does to go up), the ball would be in the air for about 6.12 seconds. To find the vertical distance the ball will travel before hitting the ground, we can use the equation for displacement s = ut + 1/2 at2 where u is the initial velocity, a is the acceleration due to gravity, and t is the time to reach the peak. Therefore, s = 30 m/s × 3.06 s + 1/2 × (-9.8 m/s2) × (3.06 s)2. After calculating, we find that s is approximately 45 m. Since the ball travels upwards and then downwards the same distance. Thus, the correct answer is D. 50 m, which is the closest rounded value. So, the ball will travel approximately 50 m vertically before hitting the ground.

Answer 2

First, let's think in the vertical problem:

The acceleration will be the gravitational acceleration:

g = 9.8 m/s^2

a = -9.8 m/s^2

For the velocity, we integrate over time:

v(t) = (-9.8 m/s^2)*t + v0

Where v0 is the initial velocity, in this case v0 = 30m/s.

v(t) = (-9.8 m/s^2)*t + 30m/s

Now, for the position we integrate again over time, and get:

P(t) = (1/2)*(-9.8 m/s^2)*t^2 + 30m/s*t + p0

Where p0 is the initial position, as the ball is launched from the ground, we can use p0 = 0m

p(t) = (-4.9m/s^2)*t^2 + 30m/s*t

Now, the maximum vertical height is reached when:

v(t) = 0m/s = -9.8m/s^2*t + 30m/s

t = 30m/s/9.8m/s^2 = 3.06s

Now we can evaluate the vertical position in t = 3.06s

p(3.06s) = (-4.9m/s^2)*(3.06)^2 + 30m/s*3.06 = 62m

So, rounding down, the correct option is: C. 60 m