A stone is thrown upward from a bridge at a speed of 10 m/s. It narrowly misses the bridge on the way back down, and hits the wa
2 answers:
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1) Initial velocity: 11.7 m/s
The motion of the rubber band is a projectile motion, so it has:
- A uniform motion along the horizontal direction
- A free-fall motion along the vertical direction
Since the motion along the horizontal direction is uniform, the horizontal velocity is constant, and it can be calculated as:
where
d = 12 m is the horizontal distance travelled
t = 1.60 s is the total time of flight
Substituting,
We also know that the horizontal velocity is related to the initial velocity of the projectile by
where
is the angle of projection
u is the initial velocity
Solving for u,
2) Highest point: 4.1 m
The initial velocity along the vertical direction is:
The motion along the vertical direction is a free fall motion, so we can use the following suvat equation
where
is the vertical velocity after the projectile has covered a vertical displacement of s
is the acceleration of gravity
At the point of maximum height, the vertical velocity is zero:
So, if we substitute it into the equation, we can find s, the maximum height: