How high above the beach was the ball when it was thrown?

Question

Bob has just finished climbing a sheer cliff above a level beach and wants to figure out how high he climbed. All he has to use is a baseball, a stopwatch, and a friend on the ground below with a long measuring tape. Bob is a pitcher and knows that the fastest he can throw the ball is about V_o =34.5 m/s. Bob starts the stopwatch as he throws the ball, with no way to measure the ball's initial trajectory, and watches carefully. The ball rises and then falls, and after t₁=0.910 s the ball is once again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measures that the ball hit the ground x =124 m from the base of the cliff

How high above the beach was the ball when it was thrown?

Answer 1

The velocity at which the ball was thrown (34.5 m/s) can be broken down into a horizontal component initial velocity v_ix and a vertical component initial velocity v_iy such that (v_ix)² + (v_iy)² = (34.5 m/s)². The path of the ball will be the shape of a parabola from the time it is thrown (t = 0) to the time it reaches Bob’s level (t₁ = 0.910 s). This means that the ball’s velocity in the vertical direction will be zero at the halfway point (t = 0.910s÷2). This information allows us to solve for the v_iy using the equation v_t = v₀ + at, where a = -9.81 m/s²:

0 m/s = v_iy + (-9.81 m/s²)(0.910 s÷2)

v_iy = 4.46 m/s

Now, we can use the equation (v_f)² = (v_i)² + 2ad to solve for the maximum height the ball reached above Bob:

(0 m/s)² = (4.46 m/s)² + 2(-9.81 m/s²)d

d = -(4.46 m/s)²/(2 x -9.81 m/s^-2) = 1.01 m

We can use the Pythagorean theorem to calculate v_ix:

(v_ix)² + (4.46 m/s)² = (34.5 m/s)²

v_ix = 34.2 m/s

There is no mention of air resistance or any forces that would cause the ball to accelerate in the horizontal direction; therefore, we can assume that the velocity of the ball in the horizontal direction remains constant the entire time the ball is in the air at 34.2 m/s. We can now solve for the time the ball is in the air t:

124 m = (34.2 m/s)t

t = 3.63 s

As we figured out earlier, the ball reached a maximum height at t = (0.910÷2) = 0.455 s. Therefore, the amount of time the ball spent descending toward earth from its maximum height is equal to the amount of time it was in the air minus the amount of time it took to reach maximum height:

t = 3.63 s - 0.455 s = 3.18 s (using correct significant figures).

Now, we can calculate v_y the instant before the ball hits the ground using the equation v_t = v_i + at:

v_fy = (0 m/s) + (-9.81 m/s²)(3.18 s)

= - 31.2 m/s

Finally, we can use the equation (v_f)² = (v_i)² + 2ad to solve for the vertical distance traveled by the ball from the time of maximum height to the time it hits the ground:

(-31.2 m/s)² = (0 m/s)² + 2(-9.81 m/s²)d

d = (-31.2 m/s)²÷(-2 × 9.81 m/s²) = 49.6 m

We already determined that the maximum height of the ball was 1.01 m above the height at which the ball was thrown. Therefore, the ball was thrown at a height of 49.6 m - 1.01 m = 48.6 m above the beach.

Answer 2

Answer 3

To find out how high above the beach the ball was when it was thrown, we can break down the problem into two parts: the vertical motion and the horizontal motion of the ball.

Vertical Motion:

We'll use the following kinematic equation to determine the initial height (h) above the beach:

h = (1/2) * g * t1^2

Where:

h is the initial height (what we want to find).

g is the acceleration due to gravity (approximately 9.81 m/s²).

t1 is the time it takes for the ball to reach its highest point, which is 0.910 seconds.

Now, plug in the values:

h = (1/2) * 9.81 m/s² * (0.910 s)^2

h = (1/2) * 9.81 m/s² * 0.8281 s²

h ≈ 4.0664 meters

So, the initial height above the beach is approximately 4.0664 meters.

Horizontal Motion:

We'll use the following equation to determine how far the ball travels horizontally:

x = V0x * t

Where:

x is the horizontal distance (124 meters).

V0x is the horizontal component of the initial velocity (which is the same as the initial velocity since there is no horizontal acceleration).

t is the total time of flight.

We need to find the total time of flight, which is twice the time it took for the ball to reach its highest point (since the ball spends an equal amount of time going up and coming down):

t = 2 * t1

t = 2 * 0.910 s

t = 1.82 seconds

Now, we can find V0x (the horizontal component of the initial velocity):

V0x = x / t

V0x = 124 m / 1.82 s

V0x ≈ 68.1319 m/s

Now that we have V0x, we can calculate the initial height (h) above the beach using the horizontal motion time (t) and the horizontal component of the initial velocity (V0x):

h = V0x * t - (1/2) * g * t^2

h = 68.1319 m/s * 1.82 s - (1/2) * 9.81 m/s² * (1.82 s)^2

h ≈ 123.93 meters

So, the ball was approximately 123.93 meters above the beach when it was thrown.