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 7.5 m. How high would this ball go if it were thrown straight upward at speed v0?
To find out how high the ball would go if it were thrown straight upward at the same initial speed, we need to understand the projectile motion of the ball.
When the ball is thrown upward at an angle, it follows a curved path known as a projectile motion. This motion has both horizontal and vertical components.
In this case, the initial speed of the ball is v0, and it is thrown at an angle of 59.0 degrees above the horizontal. The ball reaches a maximum height of 7.5 meters.
To find how high the ball would go if thrown straight upward, we need to analyze the vertical component of the motion. When the ball is thrown straight upward, the angle of projection relative to the horizontal is 90 degrees, making the horizontal component of the motion zero. Therefore, the vertical motion can be treated independently.
We can use the equations of motion for vertical motion to find the height the ball would reach. The key equation for vertical motion without considering air resistance is:
h = v0^2*sin^2(theta) / (2*g)
Where:
h is the height
v0 is the initial speed
theta is the angle of projection
g is the acceleration due to gravity (approximately 9.8 m/s²)
Since the ball is thrown straight upward, the angle of projection (theta) is 90 degrees. Plugging in the given information:
h = v0^2*sin^2(90) / (2*g)
The sine of 90 degrees is 1, so the equation simplifies to:
h = v0^2 / (2*g)
Substituting the value of g (9.8 m/s²), we get:
h = v0^2 / (2*9.8)
Since we have the maximum height in this scenario (7.5 meters), we can set up the equation:
7.5 = v0^2 / (2*9.8)
Rearranging the equation to solve for v0^2:
v0^2 = 7.5 * 2 * 9.8
v0^2 = 147
Taking the square root of both sides:
v0 ≈ √147
v0 ≈ 12.12 m/s
Now, we can plug this value back into the equation to find the height when the ball is thrown straight upward:
h = (12.12)^2 / (2*9.8)
h ≈ 7.63 meters
Therefore, if the ball were thrown straight upward at the same initial speed (approximately 12.12 m/s), it would reach a height of approximately 7.63 meters.