Hello! Please answer my question that I just posted I really need answers. I don’t know sorry hope u fine someone that does know
consider the motion in Y-direction
v₀ = initial velocity = 29 Sin62 = 25.6 m/s
a = acceleration = - 9.8 m/s²
t = time of travel
Y = vertical displacement = - 0.89 m
using the equation
Y = v₀ t + (0.5) a t²
- 0.89 = (25.6) t + (0.5) (- 9.8) t²
t = 5.3 sec
consider the motion along the horizontal direction :
v₀ = initial velocity = 29 Cos62 = 13.6 m/s
a = acceleration = 0 m/s²
t = time of travel = 5.3 sec
X = horizontal displacement =?
using the equation
X = v₀ t + (0.5) a t²
X = (13.6) (5.3) + (0.5) (0) t²
X = 72.1 m
d = distance traveled by the center fielder to catch the ball = 107 - x = 107 - 72.1 = 34.9 m
t = time taken = 5.3 sec
v = speed of center fielder
using the equation
v = d/t
v = 34.9/5.3
v = 6.6 m/s
consider the motion of the tennis ball in downward direction
Y = vertical displacement = 400 m
a = acceleration = acceleration due to gravity = 9.8 m/s²
v₀ = initial velocity of the ball at the top of building = 10 m/s
v = final velocity of the ball when it hits the ground = ?
using the kinematics equation
v² = v²₀ + 2 a Y
inserting the values
v² = 10² + 2 (9.8) (400)
v = 89.11 m/s
