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?
h = ho + (V^2-Vo^2)/2g
h = 1.75 + (0-19^2)/19.6 = 20.17 m.
Xo = 8 m/s = Hor. component of initial
velocity.
Yo = Ver. component of initial velocity.
Vo = 19 m/s = Total initial velocity.
cos A = Xo/Vo = 8/19 = 0.42105
A = 65.1o = Angle at which rock is thrown.
Yo = Vo*sin A = 19*sin 65.1 = 17.23 m/s.
h = 1.75 + (0-17.23^2)/-19.6 = 16.9 m.
To solve the first question, we need to use the principles of projectile motion. When the rock is thrown straight up, it will reach its maximum height before falling back down due to gravity.
To find the maximum height, we can use the equation:
v^2 = u^2 - 2*g*h
Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (19.0 m/s)
g = acceleration due to gravity (9.8 m/s^2)
h = height above the ground
Rearranging the equation, we get:
h = (u^2)/(2*g)
Plugging in the values, we can calculate the height:
h = (19.0^2)/(2*9.8)
h = 361.9/19.6
h = 18.45 m
Therefore, the rock will travel approximately 18.45 meters above the ground when thrown straight up.
For the second question, when the rock is thrown in an arc, we need to calculate the maximum height above the water it will travel before falling back down.
To solve this, we can use the principle of conservation of energy. At the maximum height, all of the initial kinetic energy will be converted to potential energy.
The initial kinetic energy can be calculated using the equation:
KE = (1/2)*m*v^2
Where:
m = mass of the rock (75.0 g = 0.075 kg)
v = velocity at the maximum height (8.00 m/s)
Plugging in the values, we get:
KE = (1/2)*0.075*(8.00^2) = 2.40 J
At the maximum height, the potential energy (PE) will be equal to the initial kinetic energy (KE):
PE = KE = m*g*h
Where:
g = acceleration due to gravity (9.8 m/s^2)
h = height above the water
Rearranging the equation, we get:
h = KE / (m*g)
Plugging in the values, we can calculate the height:
h = 2.40 / (0.075 * 9.8)
h = 3.22 / 0.735
h = 4.39 m
Therefore, the rock will travel approximately 4.39 meters above the water when thrown in an arc with a speed of 8.00 m/s at its maximum height.