A batted baseball leaves the bat at an angle of 33.0∘ above the horizontal and is caught by an outfielder 380ft from home plate at the same height from which it left the bat.
1. What was the initial speed of the ball?
2. How high does the ball rise above the point where it struck the bat?
A snowball rolls off a barn roof that slopes downward at an angle of α= 44.0∘ . (See the figure below) The edge of the roof is H= 17.0m above the ground, and the snowball has a speed of v = 7.00m/s as it rolls off the roof. Ignore air resistance
1. How far from the edge of the barn does the snowball strike the ground if it doesn't strike anything else while falling?
To answer these questions, we can use the principles of projectile motion. Projectile motion is the motion of an object that is launched into the air and moves along a curved path under the influence of gravity.
1. To find the initial speed of the ball, we can use the horizontal and vertical components of its velocity. The horizontal component remains constant throughout the flight, while the vertical component changes due to the effect of gravity.
First, let's find the horizontal component of the initial velocity.
Vx = V * cos(θ)
where Vx is the horizontal component, V is the initial speed, and θ is the angle above the horizontal.
Vx = V * cos(33.0∘)
Now, let's find the vertical component of the initial velocity.
Vy = V * sin(θ)
where Vy is the vertical component.
Vy = V * sin(33.0∘)
Since the ball is caught by the outfielder at the same height from which it left the bat, we know that the vertical displacement (Δy) is zero.
Using the kinematic equation:
Δy = Vy * t + (1/2) * g * t²
where Δy is the vertical displacement, t is the time of flight, and g is the acceleration due to gravity.
Since Δy is zero and the ball reaches its maximum height during the flight, we can set Vy * t + (1/2) * g * t² to zero.
0 = V * sin(33.0∘) * t + (1/2) * 9.8 m/s^2 * t²
Solving this equation will give us the time of flight (t).
Once we have the time of flight, we can use it to find the horizontal range (R), which is the distance traveled by the ball.
R = Vx * t
Given that the horizontal range is 380 ft, we can solve this equation to find the initial speed (V).
2. To find the height the ball rises above the point where it struck the bat, we can use the same kinematic equation:
Δy = Vy * t + (1/2) * g * t²
You can use the time of flight (t) and the vertical component of the initial velocity (Vy) that we calculated earlier to find the height (Δy).
time in air:
horizontal distance=speed*time
380=Vcos33*t
t=380/Vcos33
Vertical
hf=hi+vi*t-4.9 t^2
hf, hi are zero...
t(Vsin33-4.9t)=0
Vsin33-4.9(380/Vcos33)=0
V^2=4.9*380/sin33cos33
solve for V
how high, now solve for t/2,
h=vi*t-1/2 g t^2 where t is half the time.