Question

question 15 (1 point) megan rolls a ball down a ramp until the ball hits the wall. the ramp is 2.5 m long and the distance from the end of the ramp to the wall is 1.5 m. if it took 0.5 seconds for the ball to hit the wall then what is the ball’s speed?

a \t8m/s
b \t3m/s
c \t4m/s
d \t5m/s

Explanation:

Step1: Determine total distance

The ball travels down the ramp (2.5 m) and then from the end of the ramp to the wall (1.5 m). So total distance \( d = 2.5 + 1.5 = 4 \) m.

Step2: Use speed formula

Speed \( v=\frac{d}{t} \), where \( d = 4 \) m and \( t = 0.5 \) s. So \( v=\frac{4}{0.5}=8 \) m/s? Wait, no, wait. Wait, maybe I misread. Wait, no, wait the problem: "the ramp is 2.5 m long and the distance from the end of the ramp to the wall is 1.5 m". Wait, but when the ball rolls down the ramp, then from end of ramp to wall. So total distance is 2.5 + 1.5 = 4 m. Time is 0.5 s. So speed is distance over time: \( v=\frac{4}{0.5}=8 \)? But wait, the options have 8m/s as option a. Wait, but let me check again. Wait, maybe the ramp is the path? Wait, no, the ramp is 2.5 m long, then from end of ramp to wall is 1.5 m. So total distance traveled by the ball is 2.5 + 1.5 = 4 m. Time is 0.5 s. So speed \( v = \frac{4}{0.5} = 8 \) m/s. So option a.

Wait, but let me re-express:

Step1: Calculate total distance

The ball moves along the ramp (2.5 m) and then to the wall (1.5 m). So total distance \( d = 2.5 + 1.5 = 4 \) meters.

Step2: Apply speed formula

Speed \( v = \frac{\text{distance}}{\text{time}} \). Substituting \( d = 4 \) m and \( t = 0.5 \) s, we get \( v = \frac{4}{0.5} = 8 \) m/s.

Answer:

a. 8m/s