A ball's projected at an angle X to the horizontal from a cliff 96 metres high and strikes the ground at a distance of 260 metres away from the cliff.The time of its flight is 7 seconds.If it's assumed that air resistance's negligible,calculate: 1.the initial vertical velocity of the ball, 2.the time to reach the greatest height, 3.the initial horizontal velocity, 4.the angle X,and, 5.the velocity of the projectile on hitting the ground.

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To solve this problem, we can use the equations of motion for projectile motion. Let's break down each question step-by-step:

1. To find the initial vertical velocity of the ball, we can use the formula:

v = u + at

Since the final vertical velocity (v) is 0 when the ball hits the ground, we can rearrange the formula to solve for the initial vertical velocity (u):

u = v - at

Here, v = 0 (since it hits the ground) and a = -9.8 m/s² (acceleration due to gravity). We know that the time of flight (t) is 7 seconds. Let's substitute the values and calculate:

u = 0 - (-9.8 * 7)
= 68.6 m/s

Therefore, the initial vertical velocity of the ball is 68.6 m/s.

2. To find the time it takes to reach the greatest height, we can divide the total time of flight (7 seconds) by 2, since the time taken to reach the maximum height equals half the time of flight:

Time to reach the greatest height = 7 seconds / 2
= 3.5 seconds

Therefore, it takes 3.5 seconds to reach the greatest height.

3. The initial horizontal velocity (u) remains constant throughout the projectile's motion. We can use the formula:

S = ut

We know that the horizontal distance traveled (S) is 260 m. We already have the time of flight (t) as 7 seconds. Let's rearrange the formula to solve for the initial horizontal velocity (u):

u = S / t
= 260 / 7
= 37.1429 m/s (rounded to 4 decimal places)

Therefore, the initial horizontal velocity of the ball is approximately 37.1429 m/s.

4. The angle (X) can be found using the formula:

tan(X) = (u_y / u_x)

where u_y is the initial vertical velocity and u_x is the initial horizontal velocity. Let's substitute the values and solve for X:

tan(X) = (68.6 / 37.1429)
X = tan^(-1)(68.6 / 37.1429)
≈ 60.4 degrees (rounded to 1 decimal place)

Therefore, the angle X is approximately 60.4 degrees.

5. The velocity of the projectile when it hits the ground can be calculated using the formula:

v = √(u_x^2 + u_y^2)

where u_x is the initial horizontal velocity and u_y is the initial vertical velocity. Let's substitute the values and calculate:

v = √(37.1429^2 + 68.6^2)
≈ √(1379.558 + 4703.96)
≈ √(6083.518)
≈ 78 m/s (rounded to 1 decimal place)

Therefore, the velocity of the projectile when it hits the ground is approximately 78 m/s.

To solve this problem, we can break it down into several steps:

Step 1: Find the initial vertical velocity of the ball (Vy0).
Since the ball is projected at an angle X to the horizontal, we can split the initial velocity into its vertical and horizontal components.
The vertical displacement of the ball is given as 96 meters, and the time of flight is given as 7 seconds.
We can use the formula: Vy = Vy0 + gt, where g is the acceleration due to gravity (approximately -9.8 m/s^2).
Using this formula, we can solve for Vy0:

96 = Vy0 * 7 + (1/2) * (-9.8) * 7^2

This equation simplifies to:

96 = 7Vy0 - 240.1

Rearranging the equation, we get:

7Vy0 = 96 + 240.1
7Vy0 = 336.1
Vy0 = 336.1 / 7

Therefore, the initial vertical velocity of the ball (Vy0) is approximately 48.01 m/s.

Step 2: Find the time to reach the greatest height (t_g).
The time to reach the greatest height is half of the total time of flight.
So, t_g = 7 / 2 = 3.5 seconds.

Step 3: Find the initial horizontal velocity of the ball (Vx0).
To calculate Vx0, we need the horizontal displacement and the time of flight.
The horizontal displacement is given as 260 meters, and the time of flight is 7 seconds.
We can use the formula: Vx = Vx0 + gt, where Vx is the horizontal velocity component, and g is the acceleration due to gravity (approximately -9.8 m/s^2).
Since there is no acceleration in the horizontal direction, Vx0 remains constant throughout the motion.
Hence, Vx0 = (260 meters) / (7 seconds).

Therefore, the initial horizontal velocity of the ball (Vx0) is approximately 37.14 m/s.

Step 4: Find the angle X.
To find the angle X, we can use the following formula: tan(X) = Vy0 / Vx0.
Using the values we found earlier:

tan(X) = (48.01 m/s) / (37.14 m/s).

Now, we can take the inverse tangent (tan^-1) of both sides of the equation to find X:

X = tan^-1((48.01 m/s) / (37.14 m/s)).
X ≈ 53.19 degrees.

Therefore, the angle X is approximately 53.19 degrees.

Step 5: Find the velocity of the projectile on hitting the ground.
To find the velocity of the projectile on hitting the ground, we can use the Pythagorean theorem:
V = √(Vx0^2 + Vy0^2).

Using the values we found earlier:

V = √((37.14 m/s)^2 + (48.01 m/s)^2).

Therefore, the velocity of the projectile on hitting the ground (V) is approximately 60.87 m/s.