A football player throws a ball with a horizontal distance of 31.0 m so that the other player catches it at the exact height it was initially thrown. If the ball rises and falls by 1.8 m in its trajectory, what is the initial velocity of the ball? Please give the speed and angle!
how much initial speed up, Vi, to go up 1.8 m?
m g h = (1/2)m Vi^2
Vi^2 = 2 g h
Vi^2 = 2 (9.81)(1.8)
Vi = 5.94 m/s vertical component
Now work on horizontal problem
average speed up =5.94/2 = 2.97 m/s
so time rising up = 1.8/2.97 =.606 s
it goes distance 31/2 = 15.5 m while rising
so u, the constant horizontal component of velocity = 15.5/.06 = 25.6 m/s
so in the end tinital velocity is u = 25.6 and v = 5.94
speed = sqrt(u^2+Vi^2) = 26.3 m/s
tan elevation angle = 5.94/25.6
so angle = 13.1 degrees up from horizontal
To find the initial velocity of the ball, we can use the kinematic equations of motion.
1. First, let's break down the motion of the ball. We know that the horizontal distance traveled by the ball is 31.0 m, and the vertical displacement is 1.8 m (both rising and falling).
2. We can use the equation for horizontal distance:
Horizontal distance (d) = Initial velocity (v0) * Total time in the air (t)
We can assume the total time in the air is the same for the horizontal and vertical components of the motion.
3. Next, let's find the total time in the air. Since the ball rises and falls by 1.8 m, the total vertical displacement is 3.6 m (1.8 m up + 1.8 m down). We can use the equation for vertical displacement:
Vertical displacement (dy) = (1/2) * Acceleration due to gravity (g) * Total time in the air (t)^2
Plugging in the values, we get:
3.6 m = (1/2) * 9.8 m/s^2 * t^2
Solving for t, we get:
t = sqrt(3.6 m * 2 / 9.8 m/s^2) = 0.849 s
4. Now that we have the total time in the air (t), we can find the initial velocity (v0) using the equation for horizontal distance:
31.0 m = v0 * 0.849 s
Solving for v0, we get:
v0 = 31.0 m / 0.849 s = 36.6 m/s
5. We now have the initial velocity of the ball, which is 36.6 m/s.
As for the angle at which the ball is thrown, we can use the equation for the vertical component of velocity (vy) at any given time (t):
vy = g * t
Since the ball is caught at the same height it was initially thrown, the vertical component of velocity at the end of the motion is zero. So:
0 = 9.8 m/s^2 * t
Solving for t, we get:
t = 0
This means that the vertical component of velocity is zero at the peak of the trajectory, which is half the total time. Therefore, the angle at which the ball is thrown is 45 degrees (as the initial velocity has equal horizontal and vertical components when they are the same).