Cannon Fire! Senior Physics Research Project

[Michael17]

Cannon Fire!

Can anyone please help me figre this one out;

A senior physics class conducting a research project on projectile motion constructs a device that can launch a cricket ball. The launching device is designed so that the ball can be launced at ground level with an initial velocity of 28m/s at an angle of 30 degrees to the horizontal. At what time will the ball reach its maximum height?

 

[LowlyPion]

Figure the vertical velocity component from the angle.

Divide by the magnitude of gravity that is acting to slow it down.

That's your time.

 

[thomate1]

Based on the given information, we can use the equations of projectile motion to determine the time at which the ball will reach its maximum height. The first step would be to break the initial velocity into its horizontal and vertical components. The horizontal component remains constant at 28m/s, while the vertical component can be found using trigonometry as 28m/s * sin(30) = 14m/s.

Next, we can use the equation for maximum height in projectile motion, which states that the maximum height (h) is equal to the initial vertical velocity squared (v0^2) divided by twice the acceleration due to gravity (2g). In this case, the acceleration due to gravity is -9.8m/s^2 since it acts downwards. Plugging in the values, we get:

h = (14m/s)^2 / (2 * -9.8m/s^2)
h = 98m / -19.6
h = -5m

Since the maximum height is negative, this means that the ball will reach its maximum height at 5 meters below the initial height. To find the time at which this occurs, we can use the equation for vertical displacement in projectile motion, which states that the vertical displacement (y) is equal to the initial vertical velocity (v0) multiplied by time (t) plus half the acceleration due to gravity (0.5gt^2). In this case, we know that the vertical displacement is -5m, the initial vertical velocity is 14m/s, and the acceleration due to gravity is -9.8m/s^2. Plugging in the values, we get:

-5m = (14m/s) * t + 0.5 * (-9.8m/s^2) * t^2
-5m = 14m/s * t - 4.9m/s^2 * t^2
0 = 4.9m/s^2 * t^2 - 14m/s * t - 5m

Using the quadratic formula, we can solve for t and get two solutions: t = 0.42s and t = 2.38s. Since the ball reaches its maximum height at t = 0.42s, we can conclude that the ball will reach its maximum height in approximately 0.42 seconds after being launched. However, it should be noted that this