A baseball is thrown at an angle of 23◦ relative
to the ground at a speed of 22.3 m/s. The ball
is caught 36.4649 m from the thrower.
The acceleration due to gravity is
9.81 m/s2 .
How long is it in the air?
To model a spacecraft, a toy rocket engine is securely
fastened to a large puck, which can glide with negligible
friction over a horizontal surface, taken as the xy plane.
The 4.00-kg puck has a velocity of 300
ˆ
i m/s at one instant.
Eight seconds later, its velocity is to be (800
ˆ
i $ 10.0
Assuming the rocket engine exerts a constant horizontal
force, find (a) the components of the force and (b) its
magnitude.
To find the time the baseball is in the air, we can use the kinematic equation:
y = y₀ + v₀yt + (1/2)gt²
Where:
y = displacement in the y-direction (vertical distance covered by the ball)
y₀ = initial displacement in the y-direction (height from the ground at which the ball is thrown or caught)
v₀y = initial vertical component of velocity
g = acceleration due to gravity (in this case, -9.81 m/s² since it acts downwards)
t = time
Given that the ball is thrown at an angle of 23 degrees relative to the ground, we need to calculate the initial vertical component of velocity (v₀y) using trigonometry:
v₀y = v₀ * sin(θ)
where:
v₀ = initial speed of the ball
θ = angle of projection
Now, let's plug in the values:
v₀ = 22.3 m/s
θ = 23°
v₀y = 22.3 m/s * sin(23°)
Calculating v₀y:
v₀y = 22.3 m/s * 0.3907
v₀y ≈ 8.70 m/s
Next, we need to determine the final displacement in the y-direction, y. It is given as 0 since the ball returns to the same height it was thrown from:
y = 0
Now we can rearrange the equation to solve for time (t):
t = [2(y - y₀)] / g
Given:
y₀ = 0 (since the ball starts and ends at the same height)
y = 36.4649 m
g = 9.81 m/s²
t = [2 * (36.4649 m - 0)] / 9.81 m/s²
t = (2 * 36.4649 m) / 9.81 m/s²
t ≈ 7.47 s
Therefore, the ball is in the air for approximately 7.47 seconds.
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