Question
A ball is fired with an inital velocity of 500 m/s at an angle of 30 degrees above horizontal across a level field. Calculate the maximum height, the time of flight, and the range of the ball.
Okay, so I know that this is a projectile problem but I am kind of having trouble with the givens... I set up my x and y chart and even added the y1/2 so that i can find the maximum height. But I don't really understand what the 30 degrees will help me with. Is it to find the range?
Okay, so I know that this is a projectile problem but I am kind of having trouble with the givens... I set up my x and y chart and even added the y1/2 so that i can find the maximum height. But I don't really understand what the 30 degrees will help me with. Is it to find the range?
Answers
The range depends upon the launch angle A as well as the initial velocity; that is why they have to specify what it is.
I assume you already know that the maximum height is H = Vo^2 sin^2A/(2g) and that the time of flight is
T = 2 Vo sin A/g
The range R is the time of flight times the horizontal velocity component:
R = 2 Vo^2 sin A cosA/g
= Vo^2 sin(2A)/g
I assume you already know that the maximum height is H = Vo^2 sin^2A/(2g) and that the time of flight is
T = 2 Vo sin A/g
The range R is the time of flight times the horizontal velocity component:
R = 2 Vo^2 sin A cosA/g
= Vo^2 sin(2A)/g
For the height you can use these two formulas:
t=(Vi sin A)/g
H=(Vi sin A)t - (g/2)t^2
t=(Vi sin A)/g
H=(Vi sin A)t - (g/2)t^2
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