A ball is thrown upward at a speed v0 at an angle of 52.0˚ above the horizontal. It reaches a maximum height of 7.3 m. How high would this ball go if it were thrown straight upward at speed v0?

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To find out how high the ball would go if it were thrown straight upward at speed v0, we need to consider the components of its initial velocity.

When the ball is thrown upward at an angle of 52.0˚ above the horizontal, its initial velocity has two components: the vertical component (Vy) and the horizontal component (Vx).

Given that the ball reaches a maximum height of 7.3 m, we can conclude that the vertical component of its initial velocity (Vy) would be responsible for this vertical displacement.

At the maximum height, the vertical component of the velocity becomes zero (Vy = 0 m/s) because the ball momentarily stops before falling back down due to gravity.

Using the formula for vertical motion, we can determine the time it takes for the ball to reach its maximum height:

Vy = V0y - gt

0 = V0y - gt

t = V0y / g

Given that the initial velocity of the ball is v0 and the angle above the horizontal is 52.0˚, we can find the vertical component of the initial velocity (Vy):

Vy = v0 * sin(θ)

where θ is the angle above the horizontal.

Now we can substitute the value of Vy in terms of v0:

t = (v0 * sin(θ)) / g

Using this value of time, we can find the maximum height reached by the ball:

h = V0y * t - (1/2) * g * t^2

Substituting the values, we have:

h = (v0 * sin(θ)) * [(v0 * sin(θ)) / g] - (1/2) * g * [(v0 * sin(θ)) / g]^2

Simplifying this equation will give us the height the ball would reach if thrown straight upwards with the same initial speed (v0).