Answer:
t = 123.59s
Explanation:
For the launch pad section:
Vf = Vo + a*t where Vo=0.
Vf = 35*25 = 875m/s
The distance traveled during the launch:
Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.
where d= 10937.5m; Vo=875m/s.
Solving for t:
t1 = -11.093s t2 = 98.59s
So, the total time of flight will be:
Answer:
Yes
Explanation:
The given parameters are;
The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)
The angle in which the fastball is hit, θ = 22°
The distance of the field = 96 m (315 ft)
The range of the projectile motion of the fastball is given by the following formula
Where;
g = The acceleration due to gravity = 9.81 m/s², we have;
Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.
