A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 15.0m/s at an angle 5.3∘ above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. How much did it clear the net by?
yes
find horizontal velocity.
find time to reach net
time=distance/horizontal velocity
Now find height at that net point time, due to the vertical velocity and time.
hf=hi+viv^t-1/2 g t^2
is that higher than the net?
To find out how much the ball cleared the net by, we first need to find the projectile's highest point (or peak height) and then calculate the difference between the peak height and the net's height.
To solve this problem, we can break the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component changes due to the effect of gravity. Let's analyze the vertical component first.
We can use the following kinematic equation to find the time it takes for the ball to reach its highest point:
y = yo + voy*t - (1/2)*g*t^2,
where:
y = final vertical position (0 m at the peak)
yo = initial vertical position (2.0 m)
voy = initial vertical velocity (calculated from the initial speed and launch angle)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken
Rearranging the equation and plugging in the values:
0 = 2.0 + voy*t - (1/2)*(-9.8)*t^2.
We can find the initial vertical velocity (voy) using trigonometry:
voy = v * sin(theta),
where:
v = initial speed (15.0 m/s)
theta = launch angle (5.3°)
Plugging in the values:
voy = 15.0 * sin(5.3°).
Now we can substitute this expression for voy back into the first equation and solve for t.
0 = 2.0 + (15.0 * sin(5.3°)) * t - (1/2)*(-9.8)*t^2.
To find t, we can rearrange the equation and apply the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a),
where:
a = -4.9,
b = (15.0 * sin(5.3°)),
c = 2.0.
Upon calculating t, we can determine the maximum height of the projectile using the equation:
ymax = yo + voy*t - (1/2)*g*t^2.
Substituting the values and calculating the maximum height:
ymax = 2.0 + (15.0 * sin(5.3°)) * t - (1/2)*(-9.8)*t^2.
Having found the maximum height, we can calculate the difference between this value and the net's height to determine how much the ball cleared the net by.