Bryan tosses a 75.0 g rock into the lake. the rock leaves hus hand at a height of 175 cm above the ground travelling at 19.0 m/s. How high above the ground will the rock travel if it us accidentally bryan tosses a 75.0 g rock into the lake. The rock leaves hus hand at a height of 175 cm above the ground travelling at 19.0 m/s.
How high above the ground will the rock travel if it is accidentally thrown straight up?
How high above the water will it travel if it is thrown in an arc and has a speed of 8.00 m/s at its maximum height?
To find the height above the ground in each scenario, we need to apply the equations of motion. Let's break down each scenario separately:
Scenario 1: Thrown straight up
In this case, we can assume that the initial vertical velocity of the rock is 19.0 m/s (upward), and the acceleration due to gravity is 9.8 m/s² (downward). We need to determine the maximum height the rock will reach.
To find the maximum height, we can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (19.0 m/s upward)
a = acceleration (-9.8 m/s² due to gravity)
s = displacement
Rearranging the equation to solve for 's', we have:
s = (v² - u²) / (2a)
Plugging in the values, we have:
s = (0² - (19.0)²) / (2 * (-9.8))
Calculating this, we find:
s = (-361) / (-19.6)
s ≈ 18.43 meters
Therefore, the rock will reach a maximum height of approximately 18.43 meters above the ground.
Scenario 2: Thrown in an arc
In this scenario, we need to determine the height above the water when the rock is traveling at its maximum height. We are given that the speed at the maximum height is 8.00 m/s.
At the maximum height of a projectile, the vertical component of velocity is 0 m/s. We can use this information to find the height above the water.
First, let's find the time it takes for the rock to reach its maximum height. We can use the equation:
v = u + at
Since at maximum height, v = 0 m/s and a = -9.8 m/s² (acceleration due to gravity), we have:
0 = 8 + (-9.8)t
Solving for t, we find:
t = 8 / 9.8
t ≈ 0.82 seconds
Now, let's find the displacement (height above the water) at t = 0.82 s. We can use the equation:
s = ut + (1/2)at²
Since the initial velocity is 8.00 m/s upward and the acceleration is -9.8 m/s², we have:
s = 8(0.82) + (1/2)(-9.8)(0.82)²
Calculating this, we find:
s ≈ 3.08 meters
Therefore, the rock will reach a height of approximately 3.08 meters above the water when thrown in an arc with a speed of 8.00 m/s at its maximum height.
Please note that these calculations assume ideal conditions and neglect air resistance.
First of all the mass of the rock has nothing to do with this problem.
if it is thrown straight up at at Vi = 19 m/s from a height of 1.75 meters:
V = Vi - g t
at top, v = 0
0 = 19-9.81 t
t = 1.94 seconds to top
h = 1.75 + 19(1.94) -4.9(1.94)^2
h = 20.2 meters high
if it is thrown at an angle and has a speed of 8 at the top.
At the top, motion is horizontal.
So
u = horizontal speed = 8 m/s the whole time.
u = 8 = 19 cos A
where A is angle up from horizontal at launch
so
A = 65 degrees launch angle
Vi = 19 sin 65 = 17.2 initial speed up
v = 17.2 - gt
t = 1.76 seconds to max height
h = 1.75 + 17.2(1.76) - 4.9(1.76^2)
h = 16.8 meters