A ball is thrown upward at a speed v0 at an angle of 59.0¢ª above the horizontal. It reaches a maximum height of 8.7 m. How high would this ball go if it were thrown straight upward at speed v0?
Yo = ver. = Vosin59 = 0.857Vo. = ver.
component of Vo.
Vf^2 = Yo^2 + 2g*d,
Yo^2 = Vf^2 - 2g*d,
Yo^2 = 0 + 19.6*8.7 = 170.52,
Yo = 13.06m/s.
Yo = 0.857Vo = 13.06m/s,
0.857Vo = 13.06,
Vo = 15.2m/s.
h = (Vf^2 - Vo^2) / 2g,
h = (0 - (15.2)^2) / -19.6 = 11.85m.
To find out how high the ball would go if it were thrown straight upward at speed v0, we need to understand the basic principles of projectile motion.
When the ball is thrown upward at an angle, it follows a curved trajectory due to the influence of both vertical and horizontal components of its initial velocity.
In this case, the ball is thrown upward at an angle of 59.0° above the horizontal. Let's break down the initial velocity into its vertical and horizontal components:
Vertical component: v₀sinθ
Horizontal component: v₀cosθ
Given that the ball reaches a maximum height of 8.7 m, the vertical component of the initial velocity, v₀sinθ, would be responsible for the upward motion of the ball.
To determine how high the ball would go if thrown straight upward, we can set the angle to 90°, which means the ball would be thrown vertically. Therefore, the horizontal component of the initial velocity, v₀cosθ, would be zero.
This implies that the entire initial velocity, v₀, is directed vertically. Hence, to calculate the maximum height the ball would attain in this scenario, we can use the same equation:
h = v₀² * sin²θ / (2 * g)
However, since the horizontal component (v₀cosθ) is zero, the angle (θ) does not have any influence in this situation. Therefore, we need to consider:
h = v₀² * sin²90° / (2 * g)
h = v₀² * 1 / (2 * g)
h = v₀² / (2 * g)
Therefore, to calculate how high the ball would go if thrown straight upward, we need to substitute the given values into the formula:
h = v₀² / (2 * g)