A driver takes off with a speed of 10m/s from a 5m high diing board at 30° above the horizontal how much later doesh she strike the water
A rescue plaNe carries a package of emergency supply to be dropped at a point that is very close to a target the plane travels with a velocity of 75m/s implies 135m above the target how far away from the target must the pilot release the package
On earth I assume so g/2 = 4.9 m/s^2
10 sin 30 = 5 m/s speed up at start = Vi
Hi = 5 meters
h = Hi + Vi t -4.9 t^2
when does h = 0 ?
0 = 5 + 5 t - 4.9 t^2
https://www.mathsisfun.com/quadratic-equation-solver.html
t = 1.64 seconds or -0.621
forget the negative. That was before you jumped.
Same deal
Hi 135
Vi = 0
u = 75 forever horizontal speed
h = 135 + 0 t - 4.9 t^2
0 = 135 -4.9 t^2
t = 5.25 seconds
x = u t = 75 * 5.25 = 394 meters
If that package explodes on landing, turn immediately after dropping, because you are also proceeding at horizontal speed u and it hits right under you if you do not turn.
To calculate how much later the driver strikes the water, we need to analyze the motion in horizontal and vertical directions separately.
1. Horizontal motion:
The motion is horizontal, so the initial horizontal velocity remains constant throughout the motion. The driver takes off at an angle of 30° above the horizontal, but this does not affect the horizontal motion. Therefore, the horizontal velocity remains 10 m/s.
2. Vertical motion:
The initial vertical velocity is zero as the driver starts from rest vertically. The acceleration due to gravity is 9.8 m/s². The initial vertical displacement is 5 m, and the vertical displacement at the time of hitting the water is 0 m.
Using the vertical motion equations, we can calculate the time it takes for the driver to hit the water.
Using the equation: vertical displacement = (initial vertical velocity × time) + (0.5 × acceleration due to gravity × time²)
0 = (0 × t) + (0.5 × 9.8 × t²)
0 = 4.9t²
Solving for t:
t = √(0 / 4.9)
t = 0 s (There are no real solutions for t)
Therefore, the driver hits the water instantly (or at the same time) as soon as she takes off from the diving board.
To answer the first question, we need to calculate the time it takes for the driver to strike the water after taking off from the diving board.
Step 1: Resolve the initial velocity into horizontal and vertical components.
The horizontal component of the velocity can be calculated using the formula:
horizontal velocity = initial velocity * cos(angle)
In this case, the initial velocity is 10 m/s and the angle is 30 degrees.
So, the horizontal velocity = 10 m/s * cos(30) ≈ 8.66 m/s
The vertical component of the velocity can be calculated using the formula:
vertical velocity = initial velocity * sin(angle)
In this case, the initial velocity is 10 m/s and the angle is 30 degrees.
So, the vertical velocity = 10 m/s * sin(30) = 5 m/s
Step 2: Calculate the time taken to reach the maximum height.
When an object is launched vertically, the time taken to reach the maximum height can be found using the formula:
time taken = vertical velocity / gravitational acceleration
In this case, the vertical velocity is 5 m/s and the acceleration due to gravity is approximately 9.8 m/s².
So, the time taken to reach the maximum height = 5 m/s / 9.8 m/s² ≈ 0.51 seconds
Step 3: Calculate the time taken to hit the water.
Since the time taken to reach the maximum height is the same as the time taken to hit the water in this scenario, the answer is approximately 0.51 seconds.
Therefore, the driver will strike the water about 0.51 seconds later.
Moving on to the second question, we need to calculate the distance from the target where the pilot needs to release the package.
Step 1: Identify the variables.
The velocity of the plane is given as 75 m/s and the vertical distance above the target is 135 m.
Step 2: Calculate the time taken to reach the target.
The time taken can be calculated using the formula:
time taken = vertical distance / vertical velocity
In this case, the vertical distance is 135 m and the vertical velocity is 75 m/s.
So, the time taken = 135 m / 75 m/s = 1.8 seconds
Step 3: Calculate the horizontal distance.
The horizontal distance can be found using the formula:
horizontal distance = horizontal velocity * time taken
Since the horizontal velocity is not given, we assume that it remains constant.
Therefore, the horizontal distance = 75 m/s * 1.8 seconds = 135 meters.
Therefore, the pilot needs to release the package approximately 135 meters away from the target.