You're on a magic bridge floating 25 meters above a lake. You through a rock at an angle of sixty degrees. It takes 5.8 seconds for the rock to reach it's highest point.
(a) With what speed did you throw the rock?
(b) How far away from the bridge horizontally did the rock land?
sixty degres measured how?
Measured to the horizontal
Throw*
Then the vertical velocity initial is Vsin60.
At the top, vertical velocity is zero.
Vtop=Vivertial-gt
so time to top= Vivertial/g=Vsin60 /g
then solve for V
Now how far: You have to find vertical time in air.
hf=hi+Vivertical*t-1/2 g t^2
0=23+Vsin60* t-1/2 g t^2
solve for t, time in air. Use the quadratic formula.
Now distance: distance= Vhoriz*timeinair
where vhoriz= Vcos60
To solve this problem, we can use the equations of motion and consider the vertical and horizontal directions separately.
(a) To find the initial speed with which the rock was thrown, we can use the equation for the vertical motion of a projectile. The time taken for the rock to reach its highest point is equal to the time taken for it to fall back down to the same vertical position.
Using the equation for vertical motion:
v = u + at
where v is the final vertical velocity, u is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken for the rock to reach its highest point.
Since the rock is at its highest point, the final vertical velocity is 0 m/s. We can substitute the values:
0 = u - 9.8(5.8)
0 = u - 56.84
Solving for u:
u = 56.84 m/s
Therefore, you threw the rock with an initial speed of 56.84 m/s.
(b) To find how far away from the bridge horizontally the rock landed, we can use the horizontal motion of a projectile. The vertical motion does not affect the horizontal motion, so we only need to consider the initial horizontal velocity and the time taken.
Using the equation for horizontal motion:
s = ut + 0.5at^2
where s is the horizontal distance, u is the initial horizontal velocity (the initial speed of the rock), a is the horizontal acceleration (which is 0 since there is no horizontal force acting on the rock), and t is the time taken for the rock to hit the ground.
Since the horizontal acceleration is 0, the equation simplifies to:
s = ut
Substituting the values:
s = 56.84(5.8)
s = 329.552 meters
Therefore, the rock landed approximately 329.552 meters away from the bridge horizontally.