a ball that was projected up at an angle of θ degrees with the horizontal has a flight time of 5.0s and covers a range of 49.0 m. find (a) the time it took to reach maximum height, (b) the horizontal and vertical components of its initial velocity, (c) its initial velocity,(d) the angle θ at which it was projected, and (e) the maximum height it reached.
vertical problem:
2.5 seconds up and 2.5 seconds down
v = Vi - 9.81 t
at top v = 0
so
Vi = 9.81*2.5 = 24.5 m/s up
h = Vi t - 4.9 t^2
= 24.5(2.5) -4.9(6.25)
= 61.3 - 30.6
= 30.6 meters high at top
horizontal problem
49 = u (5)
so u = 9.8 m/s
tan theta = Vi/u = 24.5/9.8
theta = 68.2 degrees
the horizontal and vertical component of its initial velocity of 5.0s and 49.0m
a) The time it took to reach maximum height can be determined by dividing the total flight time by 2. Therefore, the time to reach maximum height is 5.0s / 2 = 2.5s.
b) The horizontal component of the initial velocity remains constant throughout the motion and can be calculated using the formula: horizontal component = range / flight time. Thus, the horizontal component of the initial velocity is 49.0m / 5.0s = 9.8m/s.
The vertical component of the initial velocity can be determined using the formula: vertical component = (maximum height reached - 0) / time to reach maximum height. Since we don't have the maximum height yet, we'll calculate it in the following step.
c) To find the initial velocity, we can use the formula: initial velocity = square root of (horizontal component squared + vertical component squared). We already have the horizontal component (9.8m/s), and we'll calculate the vertical component in the following step.
d) The angle θ at which the ball was projected can be determined using the equation: tan(θ) = vertical component / horizontal component. We'll calculate the vertical component in the next step.
e) To find the maximum height, we can use the following equation: maximum height = (vertical component squared) / (2 × acceleration due to gravity). We'll calculate the vertical component and use the value of acceleration due to gravity (9.8m/s^2).
Let's solve step by step:
Step 1: Calculate the vertical component of the initial velocity.
Maximum height = (vertical component squared) / (2 × acceleration due to gravity)
49.0m = (vertical component squared) / (2 × 9.8m/s^2)
Vertical component squared = 49.0m × 2 × 9.8m/s^2
Vertical component squared = 960.4m^2/s^2
Vertical component = √(960.4m^2/s^2)
Vertical component = 31m/s
Step 2: Calculate the initial velocity.
Initial velocity = square root of (horizontal component squared + vertical component squared)
Initial velocity = √(9.8m/s)^2 + (31m/s)^2
Initial velocity = √(96.04m^2/s^2 + 961m^2/s^2)
Initial velocity = √1057.04m^2/s^2
Initial velocity ≈ 32.53m/s
Step 3: Calculate the angle θ.
tan(θ) = vertical component / horizontal component
tan(θ) = 31m/s / 9.8m/s
θ ≈ tan^(-1)(31m/s / 9.8m/s)
θ ≈ 73.4 degrees
Therefore, the answers are:
(a) The time it took to reach maximum height = 2.5s.
(b) The horizontal component of the initial velocity = 9.8m/s and the vertical component = 31m/s.
(c) The initial velocity ≈ 32.53m/s.
(d) The angle θ at which it was projected ≈ 73.4 degrees.
(e) The maximum height it reached ≈ 49.0m.
To solve this problem, we need to use the equations of motion for projectile motion. Let's break down each part step by step:
(a) To find the time it took to reach maximum height, we can use the fact that the time of flight for the entire trajectory is 5.0s. Since the projectile reaches its maximum height halfway through its total time of flight, we can simply divide 5.0s by 2 to get the time taken to reach maximum height, which is 2.5s.
(b) To find the horizontal and vertical components of its initial velocity, we can use the range and time of flight. The horizontal component of velocity remains constant throughout the motion because there are no horizontal forces acting on the projectile. Therefore, we can use the formula: range = horizontal velocity × flight time. Rearranging the formula, we have: horizontal velocity = range / flight time. Substituting the given values, we get: horizontal velocity = 49.0m / 5.0s = 9.8 m/s.
The vertical component of velocity changes because there is an acceleration due to gravity acting on the projectile. The vertical component of velocity can be found using the formula: vertical velocity = (change in vertical distance) / (time taken). The change in vertical distance is the maximum height, and the time taken is the time to reach maximum height. We are given that the maximum height is halfway through the total time of flight, which means the time to reach maximum height is 2.5s. Substituting the given values, we have: vertical velocity = (0 - 0) / 2.5s = 0 m/s.
Therefore, the horizontal component of the initial velocity is 9.8 m/s and the vertical component is 0 m/s.
(c) To find the initial velocity, we can use the Pythagorean theorem to combine the horizontal and vertical components of velocity. The initial velocity can be found using the formula: initial velocity = √(horizontal velocity)^2 + (vertical velocity)^2. Substituting the given values, we have: initial velocity = √(9.8 m/s)^2 + (0 m/s)^2 = 9.8 m/s.
Therefore, the initial velocity is 9.8 m/s.
(d) To find the angle θ at which it was projected, we can use the inverse tangent function. We can use the formula: θ = arctan(vertical velocity / horizontal velocity). Substituting the given values, we have: θ = arctan(0 m/s / 9.8 m/s) = arctan(0) = 0°.
Therefore, the angle θ at which it was projected is 0° (horizontally).
(e) To find the maximum height, we can use the formula for vertical displacement during projectile motion. The formula is: vertical displacement = vertical velocity × time + (1/2) × acceleration × time^2. Since the projectile reaches the same height at the start and end of its trajectory, the vertical displacement is zero. Rearranging the formula, we have: 0 = vertical velocity × time + (1/2) × acceleration × time^2. Substituting the given values, we have: 0 = 0 m/s × 2.5s + (1/2) × (-9.8 m/s^2) × (2.5s)^2. Solving for the vertical velocity, we get: vertical velocity = (1/2) × 9.8 m/s^2 × (2.5s)^2 / 2.5s = -12.25 m/s.
Therefore, the maximum height reached is equal to the absolute value of the vertical velocity, so the maximum height is 12.25 m.