A ball is thrown straight upward and returns
to the thrower’s hand after 2.90 s in the air.
A second ball is thrown at an angle of 32.0
◦
with the horizontal.
At what speed must the second ball be
thrown so that it reaches the same height as
the one thrown vertically? The acceleration
of gravity is 9.81 m/s
2
.
Answer in units of m/
First Ball:
Tr = Tf = T/2 = 2.9/2 = 1.45s = Time to
rise to max. ht.
Vf = Vo + gt,
Vo = Vf - gt,
Vo = o + 9.81*1.45 = 14.22m/s.=Initial velocity of 1st ball.
h = Vo*t + 0.5g*t^2,
h = 14.22*1.45 - 4.9*(1.45)2 = 10.32m.
Second Ball:
Yo = Vo*sin32 = 14.22m/s,
Vo = 14.22 / sin32 = 26.83m/s @ 32 deg = Inital velocity of 2nd ball.
To solve this problem, we can use the principles of projectile motion. First, let's analyze the vertical motion of the ball thrown upward.
For the vertically thrown ball:
- The time of flight is 2.90 seconds.
- The initial velocity is the velocity at the highest point, which is zero.
- The final velocity is also zero since the ball returns to the thrower's hand.
- The acceleration is the acceleration due to gravity, which is -9.81 m/s^2 since the ball is moving upwards against gravity.
Using the kinematic equation for vertical motion:
Final velocity (v) = Initial velocity (u) + (acceleration (a) * time (t))
We can rearrange this equation to solve for the initial velocity (u):
u = v - (a * t)
Substituting the given values:
u = 0 - (-9.81 * 2.90)
u = 28.389 m/s (rounded to three decimal places)
Now, let's analyze the second ball thrown at an angle of 32.0 degrees with the horizontal.
For the second ball:
- The time of flight will be the same as the vertically thrown ball, 2.90 seconds.
- The initial velocity has both horizontal and vertical components. Since we are only concerned with the vertical component, we need to find the vertical velocity (v_y). The vertical velocity can be calculated using the equation: v_y = v * sin(angle), where v is the magnitude of the velocity and angle is the angle of projection.
- The final velocity is zero at the highest point, same as the vertically thrown ball.
- The acceleration due to gravity is still -9.81 m/s^2.
Using the kinematic equation for vertical motion:
Final velocity (v) = Initial velocity (u) + (acceleration (a) * time (t))
We can rearrange this equation to solve for the initial velocity (u):
u = v - (a * t)
Substituting the given values:
u = 0 - (-9.81 * 2.90)
u = 28.389 m/s (rounded to three decimal places)
Since the initial vertical velocities for the vertically thrown ball and the second ball are the same, we can conclude that to reach the same height, the initial vertical component of the velocity for the second ball must also be 28.389 m/s.
To find the magnitude of the total velocity of the second ball, we need to consider both the horizontal and vertical components. The horizontal component of the velocity remains constant throughout the motion.
The horizontal component (v_x) can be calculated using the equation: v_x = v * cos(angle), where v is the magnitude of the velocity and angle is the angle of projection.
Since we know the angle of projection is 32.0 degrees, and we want the magnitude of the velocity, we can use the following equation to calculate the magnitude of the velocity (v):
v = v_x / cos(angle)
v = v_x / cos(32.0)
Let's substitute the vertical velocity (v_y) we calculated earlier and solve for the magnitude of the velocity (v):
v = (v_y) / cos(32.0)
v = (28.389) / cos(32.0)
Using a scientific calculator, we can calculate this value:
v ≈ 33.25 m/s (rounded to two decimal places)
Therefore, the second ball must be thrown with a speed of approximately 33.25 m/s to reach the same height as the vertically thrown ball.