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=4 m, h=1.7 m, v=2 m/s
H-h= g•t²/2
t=sqrt[2• (H-h)/g] = …
s=v•t=...
So Elena gave you the recipe, plug the numbers in. (H = 4.5 not 4 though)
t is the time falling and the friend walks s meters in t seconds.
that doesnt look like projectile motion
got the answer which was 1.52 m
thank you!!!!!!!!!!
To calculate the distance at which the student should release the water bomb, we need to consider the time it takes for the water bomb to fall and the horizontal distance covered by the moving target during that time.
First, let's calculate the time it takes for the water bomb to fall. We can use the equation for the time of fall:
h = (1/2) * g * t^2,
where h is the height from where the water bomb is released (4.5 m) and g is the acceleration due to gravity (approximately 9.8 m/s^2). We can rearrange the equation for time (t):
t^2 = (2*h) / g,
t^2 = (2 * 4.5 m) / 9.8 m/s^2,
t^2 ≈ 0.918,
t ≈ sqrt(0.918) ≈ 0.958 s.
Now, let's calculate the distance covered by the moving target during this time. We can use the equation for distance:
d = v * t,
where d is the distance covered by the target, v is the velocity of the target (2 m/s), and t is the time it takes for the water bomb to fall (0.958 s):
d = 2 m/s * 0.958 s,
d ≈ 1.916 meters.
Therefore, the student should release the water bomb when their friend is approximately 1.916 meters away from the point of impact.