1)A ball is thrown so that it initially travels at 10ms-1 at an angle of 70 above the horizontal

a)a simple model assumes that the ball is thrown from ground level and lands at the samelevel.Find the time of flightand range of the ball.

b)A refined model assumes that the ball is at a height of 2m when it is thrown and lands at ground level. Find the range of the ball based on this assumption.

pleasehelp with a nice explanation easy to understad so i get this

1. Vo = 10m/s[70o].

Xo = 10*Cos70 = 3.42 m/s.
Yo = 10*sin70 = 9.40 m/s.

a. Y = Yo + g*Tr = 0 @ max ht.
9.40 + (-9.8)Tr = 0
Tr = 0.959 s. = Rise time.
Tf = Tr = 0.959 s. = Fall time.
T = Tr+Tf = 0.959 + 0.959 = 1.92 s. = Time in flight.
Range = Xo*T = 3.42 * 1.92 = 6.56 m.

b. Tr = 0.959 m/s, Same as part a.
h = ho + Yo*Tr + 0.5g*Tr^2.
h = 2 + 9.4*0.959 - 4.9*0.959^2 = 6.51 m. above gnd.

h = 0.5g*Tf^2 = 6.51.
4.9Tf^2 = 6.51
Tf^2 = 1.33
Tf = 1.15 s. = Fall time.

Range = Xo*(Tr+Tf) = 3.42(0.959+1.15) = 7.22 m.
The increase in range is due to the increase in fall distance and fall time(Tf).

Well, well, well, if it isn't my favorite set of physics problems! Let's tackle them together, shall we?

a) In the simple model, we assume that the ball is thrown from the ground level and lands at the same level. To find the time of flight, we need to break down the initial velocity into horizontal and vertical components. Don't worry, I won't get too mathematical on you!

The initial velocity can be broken down into two parts: the vertical component and the horizontal component. The vertical component is given by 10 m/s × sin(70°), while the horizontal component is given by 10 m/s × cos(70°).

Now, let's focus on the vertical component. When the ball is at its maximum height, its vertical velocity becomes 0. We can use this information to find out how long it takes for the ball to reach its maximum height. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Here, v = 0 m/s, u = 10 m/s × sin(70°), and a = -9.8 m/s^2 (due to gravity). By solving for t, we can find the time it takes for the ball to reach its maximum height.

Once we find the time it takes to reach the maximum height, we can simply double it to get the total time of flight. This is because the time taken to reach the maximum height is equal to the time taken to descend from the maximum height to the ground.

Now, for the range of the ball, we can use the horizontal component of the initial velocity. We can calculate this horizontal distance by multiplying the horizontal component of the initial velocity by the time of flight.

b) In the refined model, we are assuming that the ball is thrown from a height of 2m and lands at ground level. To find the range, we need to consider the horizontal component of the initial velocity (like before). However, this time, we need to subtract the time it takes for the ball to reach its maximum height from the total time of flight. Why? Because the ball is thrown from a height of 2m, so it takes less time to reach the ground.

To calculate the new time of flight, we use the same method as in part (a), but we subtract the time taken to reach the maximum height from the maximum time of flight. Once we find this new time of flight, we can multiply it by the horizontal component of the initial velocity to get the range.

There you have it! A nice little explanation for you. I hope it wasn't too confusing. If you have any more questions, feel free to ask. I'm here to help (and make you laugh)!

Sure! I'd be happy to help you understand how to calculate the time of flight and range of a ball thrown at different angles.

a) In this case, we will assume that the ball is thrown from ground level and lands at the same level. To find the time of flight and range of the ball, we can break down the motion into its horizontal and vertical components.

First, let's find the time of flight. The initial vertical velocity of the ball can be calculated using the equation:

v₀y = v₀ * sin(θ)

where v₀ is the initial velocity of the ball, and θ is the angle above the horizontal.

Plugging in the given values:
v₀ = 10 ms⁻¹
θ = 70°

v₀y = 10 ms⁻¹ * sin(70°)
≈ 9.396 ms⁻¹

Next, we can find the time it takes for the ball to reach its maximum height. The vertical motion of the ball can be described by the equation:

y = v₀y * t - (1/2) * g * t²

where y is the vertical position, t is the time, and g is the acceleration due to gravity (approximately 9.8 ms⁻²).

At the highest point, the vertical velocity becomes zero, so we can set v₀y - g * t = 0 and solve for t:

t = v₀y / g

Plugging in the values:
t = 9.396 ms⁻¹ / 9.8 ms⁻²
≈ 0.958 seconds

Since the time of flight is symmetrical, it takes the same amount of time for the ball to reach the ground from its maximum height. Therefore, the total time of flight is approximately 2 * 0.958 = 1.916 seconds.

To find the range of the ball, we can use the equation:

R = v₀x * t

where v₀x is the initial horizontal velocity of the ball and t is the time of flight.

The horizontal velocity can be calculated as:

v₀x = v₀ * cos(θ)

Plugging in the given values:
v₀ = 10 ms⁻¹
θ = 70°

v₀x = 10 ms⁻¹ * cos(70°)
≈ 3.171 ms⁻¹

Finally, we can find the range:

R = 3.171 ms⁻¹ * 1.916 seconds
≈ 6.075 meters

So, in this simplified model, the time of flight is approximately 1.916 seconds, and the range is approximately 6.075 meters.

b) In this refined model, the ball is thrown from a height of 2 meters and lands at ground level. The time of flight in this case will still be the same as in part a), which is approximately 1.916 seconds.

To find the range, we need to consider the initial vertical displacement of the ball. The equation for the vertical position is:

y = y₀ + v₀y * t - (1/2) * g * t²

where y₀ is the initial vertical position.

Plugging in the values:
y₀ = 2 meters
v₀y = 9.396 ms⁻¹
t = 1.916 seconds
g = 9.8 ms⁻²

y = 2 + 9.396 * 1.916 - (1/2) * 9.8 * (1.916)²
≈ 15.703 meters

So, in this refined model, the range of the ball is approximately 15.703 meters.

Sure! Let's break down the problem into two parts:

a) Simple Model:
In a simple model, we assume that the ball is thrown from ground level and lands at the same level. To find the time of flight and range of the ball, we need to use the equations of motion.

1. Time of Flight:
The time of flight (T) is the duration for which the ball remains in the air. We can find it using the equation:
T = (2 * usinθ) / g

where u is the initial velocity of the ball (10 m/s), θ is the angle of projectile motion (70°), and g is the acceleration due to gravity (9.8 m/s²).

Substituting the values into the equation:
T = (2 * 10 * sin(70°)) / 9.8 ≈ 1.17 seconds

Therefore, the time of flight of the ball is approximately 1.17 seconds.

2. Range:
The range (R) is the horizontal distance covered by the ball. We can find it using the equation:
R = ucosθ * T

Substituting the values into the equation:
R = 10 * cos(70°) * 1.17 ≈ 3.55 meters

Therefore, the range of the ball is approximately 3.55 meters.

b) Refined Model:
In a refined model, we assume that the ball is initially at a height of 2 meters when thrown and lands at ground level. To find the range of the ball based on this assumption, we still use the equations of motion but with an additional consideration for the initial vertical displacement.

1. Total Time of Flight:
The total time of flight (T_total) is the sum of the time taken to reach the maximum height (T_up) and the time taken to descend from the maximum height to the ground (T_down).

T_total = 2 * T_up = T_down

2. Time to Reach Maximum Height:
The time taken to reach the maximum height (T_up) can be found using the equation:
T_up = u * sinθ / g

Substituting the values into the equation:
T_up = (10 * sin(70°)) / 9.8 ≈ 1.07 seconds

Therefore, the time to reach the maximum height is approximately 1.07 seconds.

3. Time to Descend:
The time taken to descend from the maximum height to the ground (T_down) can be found using the equation:
T_down = sqrt(2h / g)

where h is the initial vertical displacement (2 meters) and g is the acceleration due to gravity (9.8 m/s²).

Substituting the values into the equation:
T_down = sqrt(2 * 2 / 9.8) ≈ 0.64 seconds

Therefore, the time to descend is approximately 0.64 seconds.

4. Total Time of Flight:
T_total = 2 * T_up = T_down = 2 * 1.07 ≈ 2.14 seconds

5. Range:
To find the range (R) based on the refined model, we use the horizontal component of the initial velocity and the total time of flight:
R = u * cosθ * T_total

Substituting the values into the equation:
R = 10 * cos(70°) * 2.14 ≈ 7.43 meters

Therefore, the range of the ball based on the refined model is approximately 7.43 meters.

I hope this explanation helps you understand how to solve both parts of the problem! Let me know if you have any further questions.