Question

A rock thrown with speed 11.5m/s and launch angle 30.0?(above the horizontal) travels a horizontal distance of d = 20.0m before hitting the ground. From what height was the rock thrown? Use the value g = 9.810m/s2 for the free-fall acceleration.Find the height yi from which the rock was launched.

Answer 1

The rock was thrown from a height of 196.2 m.

Step-by-step explanation:

To find the height from which the rock was thrown, we can use the equations of projectile motion. Since the rock was thrown horizontally, the initial vertical velocity is 0 m/s.

Using the equation of motion for vertical motion: y = yi + viyt - (1/2)gt^2, where y is the final vertical displacement, yi is the initial vertical displacement, viy is the initial vertical velocity, g is the acceleration due to gravity, and t is the time of flight.

Since the rock was thrown horizontally, viy = 0 m/s.

Plugging in the known values, we have: 0 = yi + 0 * t - (1/2) * 9.81 * (t^2)

Simplifying the equation, we get: yi = (1/2) * 9.81 * (t^2)

Substituting the given time of flight, t = 20.0 m into the equation, we can solve for yi.

yi = (1/2) * 9.81 * (20.0^2) = 196.2 m

Answer 2

The rock was launched from approximately
`8.23\; {\rm m}` above the ground. Assume that the air resistance on the rock is negligible.

Step-by-step explanation:

It is given that the initial velocity
`u = 11.5\; {\rm m\cdot s^(-1)}` is at
`30^(\circ)` above the ground. Therefore:

• Initial horizontal velocity would be:
  `u_(x) = u\, \cos(30^(\circ)) = ((√(3)) / 2) \, (11.5)\; {\rm m \cdot s^(-1)} \approx 9.9593\; {\rm m\cdot s^(-1)}`.
• Initial vertical velocity would be:
• 
  `u_(y) = u\, \sin(30^(\circ)) = (1 / 2) \, (11.5)\; {\rm m \cdot s^(-1)} = 5.75\; {\rm m\cdot s^(-1)}`.

Under the assumptions, horizontal velocity would stay the same during the entire flight:
`v_(x) = u_(x) \approx 9.9593\; {\rm m\cdot s^(-1)}`.

Since the rock travelled a horizontal distance of
`x_(x) = 20.0\; {\rm m}`, the duration of the flight would be:

`\begin{aligned}t &= (x_(x))/(v_(x)) \approx \frac{20.0\; {\rm m}}{9.9593\; {\rm m\cdot s^(-1)}} \approx 2.0082\; {\rm s}\end{aligned}`.

Also under the assumptions, vertical acceleration of the rock would be
`a_(y) = (-g) = (-9.810)\; {\rm m\cdot s^(-2)}` during the entire flight. (Negative since gravitational pull points towards the ground.)

Apply the SUVAT equation to find the vertical displacement
`x_(y)` of the rock (distance between where the rock was launched and where the rock landed):

`\begin{aligned}x_(y) &= (1)/(2)\, a_(y)\, t^(2) + u_(y)\, t \\ &\approx \left((1)/(2)\, (-9.810)\, (2.0082)^(2) + (5.75)\, (2.0082)\right)\; {\rm m} \\ &\approx (-8.23)\; {\rm m}\end{aligned}`.

(Negative since the rock landed below where it was launched.)

In other words, the rock was launched from a height of approximately
`8.23\; {\rm m}`.