A projectile is fired from ground level with an initial velocity of 50m/s and an initial angle of 30degree. Assuming g= 9.8 m/s*2, find:

Question

A projectile is fired from ground level with an initial velocity of 50m/s and an initial angle of 30degree. Assuming g= 9.8 m/s*2, find:

(a) The projectile total time of flight.
(b) The maximum height attained
(c) The total horizontal traveled
(d) The final horizontal and vertical velocities just before it hits the ground

Answer 1

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Answer 2

h = 50*sin30° t - 4.9t^2

= 25t - 4.9t^2
= t(25-4.9t)

(a) h=0 at t=0 and t=5.10

(b) max height achieved at t = 25/9.8 = 2.55
h(2.55) = 31.9

(c) x = 5.10 * 50 * cos 30° = 220.8

(d) speeds same as at launch

Answer 3

Given:

Initial velocity (u) = 50 m/s
Launch angle (θ) = 30 degrees
Acceleration due to gravity (g) = 9.8 m/s^2

We can use the following kinematic equations to solve the problem:

Vertical motion:
1. Vertical displacement (𝑦) = 𝑢 𝑡sin(θ) - 0.5 g 𝑡^2
2. Vertical velocity (𝑣𝑦) = 𝑢 𝑠𝑖𝑛(θ) - 𝑔𝑡
3. Time of flight (𝑡) = 2𝑢𝑠𝑖𝑛(θ)/𝑔
4. Maximum height attained (𝚫𝒉) = (𝑢𝑠𝑖𝑛(θ))^2 / (2𝑔)

Horizontal motion:
5. Horizontal displacement (𝑥) = 𝑢 𝑡𝑐𝑜𝑠(θ)
6. Final horizontal velocity (𝑣𝑥) = 𝑢𝑐𝑜𝑠(θ)
7. Final vertical velocity (𝑣𝑦) = 𝑔𝑡 - 𝑢𝑠𝑖𝑛(θ)

Now let's calculate each part step-by-step:

(a) The projectile's total time of flight:
𝑡 = 2𝑢𝑠𝑖𝑛(θ)/𝑔
= 2(50 m/s)sin(30°) / 9.8 m/s^2
≈ 5.11 seconds

(b) The maximum height attained:
𝚫𝒉 = (𝑢𝑠𝑖𝑛(θ))^2 / (2𝑔)
= (50 m/s * sin(30°))^2 / (2 * 9.8 m/s^2)
= (25 m/s)^2 / (19.6 m/s^2)
≈ 31.89 meters

(c) The total horizontal distance traveled:
𝑥 = 𝑢 𝑡cos(θ)
= (50 m/s)cos(30°) * (5.11 s)
≈ 220.94 meters

(d) The final horizontal and vertical velocities just before it hits the ground:
Final horizontal velocity (𝑣𝑥) = 𝑢cos(θ)
= (50 m/s)cos(30°)
≈ 43.3 m/s

Final vertical velocity (𝑣𝑦) = 𝑔𝑡 - 𝑢sin(θ)
= (9.8 m/s^2)(5.11 s) - (50 m/s)sin(30°)
≈ -24.36 m/s (negative sign indicates the downward direction)

Therefore, the final horizontal velocity is approximately 43.3 m/s and the final vertical velocity is approximately -24.36 m/s.

Answer 4

To find the various properties of the projectile, we'll need to break the initial velocity into its horizontal and vertical components.

(a) Time of Flight:

The time of flight is the total time it takes for the projectile to reach the ground again. It can be calculated using the vertical motion equation:

Vertical motion equation:
y = u * sin(theta) * t - (g * t^2) / 2

where:
y is the vertical displacement (which is zero at the top of the trajectory),
u is the initial vertical velocity (u = initial velocity * sin(theta)),
g is the acceleration due to gravity (9.8 m/s^2),
t is the time.

Since the vertical displacement is zero at the top of the trajectory, we can rewrite the equation as:
0 = (initial velocity * sin(theta)) * t - (g * t^2) / 2

Rearranging the equation, we get:
(g * t^2) / 2 = (initial velocity * sin(theta)) * t

Simplifying further:
(g * t^2) = 2 * (initial velocity * sin(theta)) * t

Dividing both sides by t:
g * t = 2 * initial velocity * sin(theta)

Dividing both sides by g:
t = (2 * initial velocity * sin(theta)) / g

Plugging in the given values:
t = (2 * 50 * sin(30)) / 9.8
t ≈ 3.06 seconds

Therefore, the projectile's total time of flight is approximately 3.06 seconds.

(b) Maximum Height:

The maximum height attained by the projectile can be determined using the vertical motion equation:

Vertical motion equation:
y = u * sin(theta) * t - (g * t^2) / 2

At the maximum height, the vertical velocity becomes zero. Therefore, we can find the time taken to reach the maximum height by setting the vertical velocity equal to zero:

0 = u * sin(theta) - g * t
u * sin(theta) = g * t

Rearranging the equation, we get:
t = (u * sin(theta)) / g

Plugging in the given values:
t = (50 * sin(30)) / 9.8
t ≈ 1.27 seconds

Using this time, we can find the maximum height by substituting it into the vertical motion equation:

y = u * sin(theta) * t - (g * t^2) / 2
y = (50 * sin(30)) * 1.27 - (9.8 * 1.27^2) / 2

Simplifying further:
y ≈ 31.8 meters

Therefore, the projectile reaches a maximum height of approximately 31.8 meters.

(c) Total Horizontal Travel:

The total horizontal displacement can be calculated using the horizontal motion equation:

Horizontal motion equation:
x = u * cos(theta) * t

Plugging in the given values:
x = (50 * cos(30)) * 3.06

Simplifying further:
x ≈ 263.2 meters

Therefore, the projectile travels approximately 263.2 meters horizontally.

(d) Final Horizontal and Vertical Velocities:

The final horizontal velocity remains constant throughout the motion and is equal to the initial horizontal velocity:

Final horizontal velocity = Initial horizontal velocity = initial velocity * cos(theta)
Final horizontal velocity = 50 * cos(30)

Simplifying further:
Final horizontal velocity ≈ 43.3 m/s

The final vertical velocity just before hitting the ground can be found using the equation for vertical motion:

Vertical motion equation:
v = u * sin(theta) - g * t

Plugging in the given values:
v = (50 * sin(30)) - (9.8 * 3.06)

Simplifying further:
v ≈ -24.4 m/s

Therefore, the final horizontal velocity just before hitting the ground is approximately 43.3 m/s, and the final vertical velocity is approximately -24.4 m/s.