Problem Solving. Anne throws a ball horizontally from the top of a building that is 50 m high. He hopes the ball will reach the swimming pool that is at the bottom of a building 12 m horizontally from the edge of the building. If the ball is to reach the pool, with what initial speed must Anne throw it with? Given: Dy = 50m Dx = 12 m Find =V1
Anne throws a ball horizontally...
so the problem reduces to a free-fall over 50m.
The equation
Dy=-50m = V0t+(1/2)gt^2
will let you solve for the time to hit ground, and where
v0=initial vertical velocity = 0, and
g=acceleration due to gravity = -9.8 m/s²
During this time, the ball must travel Dx=12m horizontally.
So using Distance = speed * time will let you solve for the horizontal speed required.
To solve this problem, we can use the principles of projectile motion.
First, we need to determine the time it takes for the ball to reach the pool horizontally. Since there is no vertical acceleration in this case, the time it takes for the ball to fall vertically (from the top of the building to the bottom) will be the same as the time it takes for the ball to travel horizontally (from the building to the edge of the building where the pool is).
We can use the equation: Dx = Vx * t, where Dx is the horizontal distance, Vx is the horizontal velocity, and t is the time.
Given that Dx = 12 m, we need to determine the time it takes for the ball to travel horizontally.
We can use the equation: Dx = Vx * t, rearrange it to solve for t: t = Dx / Vx.
Using this equation, we can substitute the given values:
t = 12 m / Vx.
Next, we need to determine the initial vertical velocity (Vy) of the ball. The ball is dropped vertically from a height of 50 m, so we can use the equation for vertical displacement to solve for Vy.
Using the equation: Dy = Vy * t + 0.5 * g * t^2, where Dy is the vertical displacement, Vy is the vertical velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s^2).
Given that Dy = 50 m, we can substitute the values and solve for Vy.
50 m = Vy * t + 0.5 * (-9.8 m/s^2) * t^2.
Now we have two equations: t = 12 m / Vx and 50 m = Vy * t + 0.5 * (-9.8 m/s^2) * t^2.
We can substitute the value of t from the first equation into the second equation:
50 m = Vy * (12 m / Vx) + 0.5 * (-9.8 m/s^2) * (12 m / Vx)^2.
Simplifying this equation will give us an equation with only one variable (Vx):
50 m = 12 m * Vy / Vx - 0.5 * 9.8 m/s^2 * (12 m / Vx)^2.
Solving this equation for Vx will give us the initial velocity needed:
Vx = (12 m * Vy) / (50 m + 0.5 * 9.8 m/s^2 * (12 m)^2 / Vx).
Finally, we can substitute the value of Vy = 0 (because the initial vertical velocity is zero when the ball is thrown horizontally) and solve for Vx:
Vx = (12 m * 0) / (50 m + 0.5 * 9.8 m/s^2 * (12 m)^2 / Vx).
Since the numerator is zero, the initial horizontal velocity (Vx) is also zero.
Therefore, Anne must throw the ball horizontally with an initial speed of zero to reach the swimming pool.