A foam dart is aimed horizontally at a target 1.97 m away. The dart hits 0.206 m below the target. Ignore air resistance and assume that g = 9.8 m/s2. What was the initial speed of the dart in meters per second?
Answers
Answer: 46.9 m/s
Explanation:
The dart is thrown in horizontal direction. Let the initial speed with which the dart is thrown be u.
The initial speed in vertical direction is 0. But the dart falls under acceleration due to gravity.
From the equation of motion, we will find out the time in which dart covers vertical displacement of s = 0.206 m.
s = u t + 0.5 a t²
⇒0.206 m=0+0.5×9.8 m/s²×t²
⇒t=√0.206÷√4.9=0.042 s
Now, in the same time, the dart would cover the horizontal distance, d=1.97 m
Speed, u = d/t = 1.97 m ÷ 0.042 s = 46.9 m/s
Hence, the initial speed of dart is 46.9 m/s in the horizontal direction
Given:
Distance of the target = 1.97 m
Distance missed from the target = 0.206 m
Gravity = 9.8 m/s
To find:
The initial speed of the dart in meters per second
Solution:
To find the initial velocity,
Initial velocity = 0 ( vertically )
By the laws of motion,
Displacement = Initial velocity * time + 0.5 * gravity * ( time )^2
s = u t + 0.5 a t^2
0.206 = 0 + 0.5 * 9.8 * t²
t = 0.042 s
Speed = distance / time
Speed = 46.9 m/s
Hence, the initial speed of dart is 46.9 meters per second.