3. Baseball pitcher Roy Halladay throws a 40 m/s fastball right down the center

Question

3. Baseball pitcher Roy Halladay throws a 40 m/s fastball right down the center on a 3-2 count to strike out A. Rod. How long did it take for the ball to reach the batter if the pitcher' s mound is 18m away from home plate? Also, how far did the ball drop in that time from the height that Halladay released it at? You can neglect air resistance. (0.45s, 0.99m) 4. An arrow leaves a bow with a speed of 42 m/s. Its velocity is reduced to 34 m/s by the time it hits its target. How much distance did the arrow travel over if it were in the air for 2.4 seconds? (-3.3m/s2, 92

Explanation / Answer

3)

Say the initial point of release was at a height H mts from the ground.
The initial velocity is u = 40m/s
Let the time taken for the ball to reach the mound is t seconds
Using kinematic equation of motion s = ut + 0.5at2 for the horizontal motion of 18mts

After substitution we get
18 = 40*t + 0
t = 0.45 seconds

For vertical motion, the acceleration is 9.81 m/ss in the downward direction that is the gravitational acceleration g
The final velocity of the ball is v'
again using v' = u +at
Here t is the above time, initial velocity u in the vertical direction is 0 m/s
v' = 0 + (-9.81)*0.45
v' = -4.41 m/s

Now using 2as = v2 - u2
s is the height through which the ball has been lowered through the motion
2*(-10)*s = (-4.41)2
Solving for s we get s = 1 mt

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.