A friend throws a baseball horizontally. He releases it at a height of 2.0 m and it lands 21 m from his front foot, which is directly below the point at which he released the baseball. (a) How long was it in the air? (b) How fast did he throw it?
First solve the vertical plane problem.
How long does it take to drop from 2 meters ? It has zero initial velocity down.
-2 = 0 + 0 t +(1/2)(-9.8) t^2
t^2 = 4/9.8
t = .64 seconds in the air
then distance = speed * time
21 = speed * .64
thankyou!!
wait i believe this is not my problem? mine is the cannon problem
Same problem, different numbers
To find the answers to the given questions, we can use the equations of motion for horizontal motion and vertical motion. Let's solve these step by step:
(a) How long was it in the air?
In the horizontal direction, there is no acceleration because there is no force acting horizontally on the baseball. So the horizontal velocity is constant throughout the motion.
We can use the equation: distance = velocity × time
Given that the baseball lands 21 m away from the starting point, and we know the velocity is constant, we can write:
Distance = Velocity × Time
Rearranging the equation to solve for time, we get:
Time = Distance / Velocity
Substituting the values given, we have:
Time = 21 m / Velocity
(b) How fast did he throw it?
In the vertical direction, the only force acting on the baseball is gravity, which causes it to accelerate downward. The initial vertical velocity is zero because the ball is thrown horizontally. The height of the ball does not affect its time of flight, only the shape of the trajectory.
We can use the equation for vertical displacement: h = (1/2)gt^2, where h is the vertical displacement, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.
Given that the initial height is 2.0 m, the vertical displacement is -2.0 m (since it lands on the ground), we can write:
-2.0 m = (1/2)(9.8 m/s^2)t^2
Simplifying the equation, we have:
t^2 = (-2.0 m) / (1/2)(9.8 m/s^2)
t^2 = -0.408163
Taking the square root of both sides, we get:
t ≈ 0.638 s
So, the time of flight is approximately 0.638 seconds.
To find the velocity at which the baseball was thrown, we can use the horizontal distance and the time of flight, as the horizontal velocity is constant.
Velocity = Distance / Time
Substituting the values given, we have:
Velocity = 21 m / 0.638 s
Velocity ≈ 32.91 m/s
Therefore, the baseball was thrown at a speed of approximately 32.91 m/s.