A projectile is launched upward with a velocity of 128

Question

A projectile is launched upward with a velocity of 128

feet per second from the top of a 50- foot platform. What is the maximum height attained by the projectile?

Answer 1

g = -32 ft/s^2

so it take 4 seconds to reach max height ... 128 / 32 = 4

the average velocity is ... (128 + 0) / 2 = 64 ft/s

max height = 50 + (64 * 4)

Answer 2

Well, if a clown were to launch themselves upward with a velocity of 128 feet per second from a 50-foot platform, they would probably hit the ceiling and make quite a mess! But let's get back to your question. To find the maximum height attained by the projectile, we can use the equation of motion. The formula for the maximum height can be written as h = (v^2)/(2g), where v is the initial velocity and g is the acceleration due to gravity. So, plugging in the values, we get h = (128^2)/(2*32.2). Doing the math, we find that the maximum height attained by the projectile is approximately 256 feet.

Answer 3

To find the maximum height attained by the projectile, we can use the kinematic equation for the vertical motion:

v^2 = u^2 - 2as

Where:
v = final velocity (0 ft/s at the maximum height)
u = initial velocity (128 ft/s)
a = acceleration due to gravity (-32 ft/s^2 for objects launched upward)
s = displacement (maximum height attained)

Rearranging the equation, we have:

s = (u^2 - v^2) / (2a)

Substituting the given values:

s = (128^2 - 0^2) / (2 * (-32))

s = 16384 / (-64)

s = -256

The calculated value for s is negative because the displacement is measured from the platform's height. Therefore, the maximum height attained by the projectile is 256 feet above the platform.

Answer 4

To find the maximum height attained by the projectile, we'll use the kinematic equation for vertical displacement:

𝑦 = 𝑦0 + 𝑣0𝑡 - 1/2𝑔𝑡²

Where:
- 𝑦 is the final vertical position (maximum height)
- 𝑦0 is the initial vertical position (the height of the platform)
- 𝑣0 is the initial velocity in the vertical direction
- 𝑔 is the acceleration due to gravity (approximately -32 feet per second squared)
- 𝑡 is the time taken

In this case, 𝑦0 is 50 feet (the height of the platform) and 𝑣0 is 128 feet per second (the initial velocity). We want to find 𝑡, the time taken to reach the maximum height.

At the maximum height, the vertical velocity becomes 0. So we can use the equation:

0 = 𝑣0 - 𝑔𝑡

Rearranging the equation, we have:

𝑡 = 𝑣0 / 𝑔

Substituting the values, we have:

𝑡 = 128 / 32 = 4 seconds

Now, we can use this time value to find the maximum height. Plugging in the values into the first equation:

𝑦 = 𝑦0 + 𝑣0𝑡 - 1/2𝑔𝑡²

𝑦 = 50 + 128(4) - 1/2(32)(4)²

𝑦 = 50 + 512 - 1/2(32)(16)

𝑦 = 50 + 512 - 256

𝑦 = 306 feet

Therefore, the maximum height attained by the projectile is 306 feet.