A ball is thrown straight upward and returns to the thrower’s hand after 2 s in the air. A second ball is thrown at an angle of 38 degrees� 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 9.8 m/s2.
Answer in units of m/s
To solve this problem, we need to analyze the motion of both balls and find the velocity of the second ball when it reaches the same height as the vertically thrown ball.
Let's start by considering the vertically thrown ball. We know that it takes 2 seconds for the ball to go up and then return to the thrower. In the first 1 second, the ball goes up, and in the next 1 second, it comes back down. Since the ball goes up and falls back down, the total time for the ball to reach the peak height is 1 second.
Using the equation of motion for vertical motion, we have:
h = ut + (1/2)gt^2
where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
For the vertically thrown ball, at the peak height, the final height will be the same as the initial height, and the final velocity will be zero. Therefore, the equation becomes:
0 = u(1) + (1/2)(-g)(1)^2
Simplifying the equation, we have:
0 = u - (1/2)g
Rearranging the equation, we get:
u = (1/2)g
Now let's consider the second ball thrown at an angle of 38 degrees with the horizontal. We need to find the initial velocity of the second ball (v) so that it reaches the same height as the vertically thrown ball.
The horizontal component of the velocity (vx) remains constant throughout the motion since there is no horizontal force acting on the ball. The vertical component of the velocity (vy) changes due to the acceleration of gravity.
To find the initial velocity (v), we can use the equation:
u = v * cos(theta)
where u is the initial velocity, v is the magnitude of the initial velocity, and theta is the angle of projection.
Substituting the known values, we have:
(1/2)g = v * cos(38)
Rearranging the equation, we get:
v = (1/2)g / cos(38)
Using the given acceleration due to gravity (g = 9.8 m/s²) and the angle in degrees, we can now calculate the velocity:
v = (1/2)(9.8 m/s²) / cos(38°)
v ≈ 5.2475 m/s
Therefore, the second ball must be thrown at a speed of approximately 5.2475 m/s to reach the same height as the vertically thrown ball.
It has to have the same vertical component of velocity
original ball:
Vf=Vi+gt
-Vi-Vi=-9.8t
Vi=4.9m/s
second ball
VsinTheta=4.9