the height from where the water bomb will be released on the overpass is 4.5 m and the target student is 170 cm tall. the student realises that, as his target will be moving, he must release the water bomb when his friend is a certain horizontal distance from the point of impact. if his friend is walking towards the overpass at 2m/s calculate the distance
h = ho - (Vo*t + 0.5g*t^2 = 1.71 m.
4.5 - (0 + 4.9t^2 = 1.71
-4.9t^2 = 1.71 - 4.5 = -2.79
t^2 = 0.5694
Tf = 0.755 = Fall time.
d = 2m/s * 0.755s = 1.51 m.
To calculate the horizontal distance, we need to consider the time it takes for the water bomb to fall from the overpass to the ground.
First, we convert the target student's height from centimeters to meters:
Target student's height = 170 cm = 170/100 = 1.7 m
Next, we need to calculate the time it takes for the water bomb to fall from a height of 4.5 m. We can use the equation for vertical distance traveled by an object in free fall:
h = (1/2) * g * t^2
Where:
h = vertical distance (4.5 m)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Substituting the values into the equation:
4.5 = (1/2) * 9.8 * t^2
Rearranging the equation to solve for t:
t^2 = (4.5 * 2) / 9.8
t^2 = 0.9184
t = sqrt(0.9184)
t ≈ 0.959 seconds
Now that we know the time it takes for the water bomb to fall, we can calculate the distance traveled by the friend during this time. The distance traveled can be found using the equation:
distance = speed * time
where:
speed = 2 m/s
time = 0.959 seconds
distance = 2 * 0.959
distance ≈ 1.918 meters
Therefore, the friend will need to be at a horizontal distance of approximately 1.918 meters from the point of impact when the water bomb is released.