Question

Q.N.3) A baseball stadium is designed so that the playing field is the region bounded by the circle is with the equation (x-104) ^2+(y-104) ^2=256^2 and the first quadrant. Home plate is located at the origin. If the ball is hit with an initial velocity of 110mphand the angle of 30⁰ elevation, describe the home run zone in terms of an acute angle, A, with vertex at home plate and which is measured from the first base line (i.e. A is 0⁰ at the first base line and 90⁰ at the third base line. Assume the fence is 10ft tall).

Answer 1

The initial velocity (u) of the baseball is 110 mph. The angle of elevation (6) is 30° the time of flight for the ball, which is in the projectile motion is, 211 sin θ Calculate the time of flight for the ball as follows 211 sin θ (2)(110mph)(sin 30) 9.8 m/s2 (16x10 m (sin 30°) (2)| 110 3600s 9.8 m/s だ4.9s The equation for the projectile ball in the horizontal direction is,

Calculate the distance travelled by the ball in the horizontal direction as follows: Here, the acute angle (A) is (0 X-(11 COSA): x= (110mph)(cos 0°)(4.9s) X- x 239.5m Calculate the distance travelled by the ball in the horizontal direction as follows: Here, the acute angle (A) is 90 1.6x103 m 3600s 11(1)(4.9 s) 110 x (110 mph) (cos 90) (4.9s) The equation for the projectile ball in the vertical direction is,

Calculate the distance travelled by the ball in the vertical direction as follows: Here, the acute angle (A) is 0 y-(10mph )(sin'9m/s49) 9.8m/s)(4.9s 10 100/a9s 1.6×103 m 3600s sin 0° (4.9s 9.8m/s2 (4.9s yー-117.6m Calculate the distance travelled by the ball in the horizontal direction as follows Here, the acute angle (A) is 90 y-(10mph)(sin90%(49)8s)(49 9.8m/s2 )(4.9s 1010904)08m/)49s 1.6x10 m 3600s У -| sin 90°) (4.9s 9.8m/s2)(4.9s) y 121.9m