A major league pitcher throws the ball at a speed of 159.2 ft/s. How high would the ball go if thrown straight up? Round the final answer to the nearest whole number.
The correct answer and explanation is:
To find how high the ball would go if thrown straight up at a speed of 159.2 ft/s, we use a kinematic equation from physics: v2=v02−2ghv^2 = v_0^2 – 2g h
Where:
- vv = final velocity (0 ft/s at the top)
- v0v_0 = initial velocity = 159.2 ft/s
- gg = acceleration due to gravity = 32.2 ft/s²
- hh = maximum height (what we’re solving for)
Step-by-step calculation:
0=(159.2)2−2(32.2)h0 = (159.2)^2 – 2(32.2)h (159.2)2=2(32.2)h(159.2)^2 = 2(32.2)h 25344.64=64.4h25344.64 = 64.4h h=25344.6464.4≈393.7 feeth = \frac{25344.64}{64.4} \approx 393.7 \text{ feet}
Final Answer: 394 feet\boxed{394 \text{ feet}}
Explanation (300 words):
This problem involves basic concepts of kinematics — specifically motion under constant acceleration. When a pitcher throws a baseball straight up, the only force acting on the ball (after release) is gravity, which slows the ball down until it momentarily stops at the highest point. At that point, the velocity becomes zero, and then the ball begins to fall back down.
We use the kinematic equation v2=v02−2ghv^2 = v_0^2 – 2gh, which is ideal for situations where you know the initial velocity, the final velocity (which is 0 ft/s at the highest point), and want to calculate displacement (height).
Here, the initial velocity v0v_0 is 159.2 feet per second — a fast pitch — and gg, the acceleration due to gravity, is taken as 32.2 ft/s², a standard value used in U.S. customary units. Plugging the numbers into the formula, we isolate and solve for hh, the maximum height.
The resulting value is about 393.7 feet. Since the problem asks for the final answer to be rounded to the nearest whole number, we round up to 394 feet.
This calculation assumes no air resistance and that the throw is perfectly vertical, making it a simplified physics problem useful for learning basic motion concepts.