Answer:
Step-by-step explanation:
We can solve this problem using kinematic equations. We know that the initial velocity of the ball is 55 m/s at an angle of 30° to the horizontal. We can break this velocity into its horizontal and vertical components:
vx = v0 cos θ = 55 cos 30° = 47.6 m/s
vy = v0 sin θ = 55 sin 30° = 27.5 m/s
We can now use the vertical motion equation to find the time it takes for the ball to reach its maximum height:
Δy = vy t + 0.5 a t^2
At the maximum height, the vertical velocity of the ball is 0, so we have:
0 = vy + a t_max
Solving for t_max, we get:
t_max = -vy / a = -27.5 / (-9.8) = 2.81 s
The ball will take twice this time to reach the fence, since it needs to come back down to the ground:
t_total = 2 t_max = 5.62 s
The horizontal distance the ball travels during this time is:
Δx = vx t_total = 47.6 × 5.62 = 267.7 m
Since this distance is greater than the distance to the fence (120 m), the ball will reach the fence if it leaves the bat traveling upward at an angle of 30° to the horizontal.