A batted baseball leaves the bat at an angle of 30.0 degrees about the horizontal and is caught by an outfilder 375 ft from home plate at the same height from which it left the bat. (a) what was the initial speed on the ball? (b) how high does the ball rise about the point where is struck the bat?
A) timeinair=375ft/Vi*.866=433/Vi
Vi*1/2=1/2 (33ft/sec^2)433/Vi
vi= sqrt (33*433) ft/sec
sqrt of 14289 ft/sec
Vi= 119.5366
The correct answer is 118 what did I do wrong
and how do I solve for B
a.
If you use g=32.2 instead of 33, you should get the right answer.
b. calculate time in air (as in a) with calculated Vi. Divide this by 2 gives the time when the ball reaches the highest point.
Use
S=Vi.sin30.t + g.t^2/2 (g is -32.2 f/s/s
Did you have time to look at the baseball problem?
The one above ??? I already solved it
Excellent!
So you are ready for the Tennis problem, or is it solved too?
To solve for part A, you correctly established the equation:
timeinair = 375 ft / (Vi * Sin30°)
However, there seems to be a mistake in your subsequent calculations. Let's go through the steps again.
From the given equation, we have:
timeinair = 375 ft / (Vi * 0.5)
Simplifying, we get:
timeinair = 750 ft / Vi
Now, we can use the kinematic equation of motion:
375 ft = (Vi * Cos30°) * timeinair
Replacing timeinair with 750 ft / Vi, we get:
375 ft = (Vi * Cos30°) * (750 ft / Vi)
Simplifying further:
375 ft = 750 ft * Cos30°
Dividing both sides by 750 ft and taking the inverse cosine of the result:
Cos30° = 375 ft / 750 ft
Cos30° = 0.5
So the initial velocity (Vi) is:
Vi = (375 ft) / (0.5 * 750 ft)
Vi = 118 ft/sec
Thus, the correct answer for part A is 118 ft/sec.
Now, let's move on to part B, which is to determine how high the ball rises. To solve this, we need to find the vertical component of the initial velocity. Given that the ball was launched at an angle of 30.0 degrees, the vertical component can be found using trigonometry.
Vertical component of initial velocity (Vyi) = Vi * Sin30°
Vyi = 118 ft/sec * Sin30°
Vyi = 118 ft/sec * 0.5
Vyi = 59 ft/sec
Now, using the kinematic equation for vertical motion:
Delta y = (Vyi^2) / (2 * g)
Where g is the acceleration due to gravity (approximately 32.2 ft/sec^2).
Delta y = (59 ft/sec)^2 / (2 * 32.2 ft/sec^2)
Delta y = 3,481 ft^2/s^2 / 64.4 ft/sec^2
Delta y ≈ 54.1 ft
Therefore, the ball rises approximately 54.1 ft above the point where it struck the bat.