Projectile fired in earths gravitational field at a height of of 0.90m above ground level, with a horizontal velocity of 9m/s. How far will the ball fly before hitting the ground? How long will the ball stay in the air?
time in air;
hf=hi -4.9timeinair^2
0=.9 -4.9 t^2 solve for timeinair t.
horizontal distance=9*t
To find the distance the ball will fly before hitting the ground and how long it will stay in the air, we can use the equations of motion for a projectile.
Let's assume that the initial height of the projectile, h, is 0.90m and the initial horizontal velocity, v, is 9m/s.
First, we can calculate the time it takes for the ball to reach the ground. We can use the equation of motion for the vertical direction:
h = (1/2) * g * t^2
where h is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes to hit the ground.
Plugging in the values, we get:
0.90 = (1/2) * 9.8 * t^2
Simplifying the equation, we get:
0.90 = 4.9 * t^2
Dividing both sides by 4.9, we get:
t^2 = 0.90 / 4.9
Taking the square root of both sides, we get:
t = √(0.90 / 4.9)
Calculating this, we find that t is approximately 0.43 seconds.
Now, let's find the horizontal distance the ball will fly in this time. We can use the equation of motion for the horizontal direction:
d = v * t
where d is the horizontal distance and v is the initial horizontal velocity.
Plugging in the values, we get:
d = 9 * 0.43
Calculating this, we find that the ball will fly approximately 3.87 meters before hitting the ground.
So, the ball will fly approximately 3.87 meters before hitting the ground, and it will stay in the air for approximately 0.43 seconds.