A ball is thrown upward at a speed v0 at an angle of 53° above the horizontal. It reaches a maximum height of 8.2 m. How high would this ball go if it were thrown straight upward at speed v0?

To determine how high the ball would go if it were thrown straight upward at speed v0, we need to consider the initial velocity and the maximum height reached in the given scenario.

When the ball is thrown upward at an angle of 53° above the horizontal, it would follow a projectile motion trajectory. The component of the initial velocity in the vertical direction determines the height the ball reaches.

Given that the maximum height reached is 8.2 m, we can use this information to find the vertical component of the initial velocity (v0y). To do this, we use the formula for maximum height in projectile motion:

Hmax = (v0y)^2 / (2 * g)

Where:
Hmax = maximum height reached
v0y = vertical component of the initial velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)

Rearranging the formula, we get:

(v0y)^2 = Hmax * 2 * g

Taking the square root of both sides, we find:

v0y = sqrt(Hmax * 2 * g)

Plugging in the values, we can now calculate v0y:

v0y = sqrt(8.2 * 2 * 9.8)
≈ 8.0 m/s

Now, if the ball were thrown straight upward, the angle above the horizontal would be 90°. This means the vertical component of the initial velocity is the same as the initial velocity itself (v0y = v0).

Therefore, in this scenario, the ball would reach a maximum height of approximately 8.0 meters, assuming the initial speed v0 remains unchanged.