karen hits a tennis ball with an initial velocity of 42 feet per second an at angle of 16 degree with the horizontal from a height of 2 feet. she is 20 feet from the net and the net is 3 feet. will the ball go over the net?
feet, yuuk :(
oh well, g = 32 ft/s^2 in that case
Vi = 42 sin 16 = 11.6 ft/s
u = 42 cos 16 = 40.4 ft/s
Horizontal problem:
goes 20 ft at 40.4 ft/s
t = 20/40.4 = .495 s
Vertical problem with t = .495 s
-m g = m d^2h/dt^2
d^2 h/dt^2 = -g = -32
dh/dt = Vi - 32.2 t
h = Hi + Vi t - (1/2)(32)t^2
h = Hi + Vi t - 16 t^2
h = 2 + 11.6(.495) - 16(.495)^2
= 3.82 feet, clears easily
To determine if the ball will go over the net, we need to calculate the trajectory of the ball and compare it to the height of the net.
To do this, we can use the equations of motion for projectile motion. The horizontal and vertical components of the ball's velocity are given by:
Vx = V * cos(theta)
Vy = V * sin(theta)
where V is the initial velocity of the ball (42 ft/s) and theta is the launch angle (16 degrees).
Next, we can calculate the time it takes for the ball to reach the net by using the equation:
t = 2 * Vy / g
where g is the acceleration due to gravity (approximately 32 ft/s^2).
Using the time, we can compute the horizontal distance the ball travels:
x = Vx * t
Finally, we can calculate the highest point reached by the ball using the equation:
y = h + Vy^2 / (2 * g)
where h is the initial height of the ball (2 feet).
If the highest point of the ball's trajectory is higher than the net's height, which is 3 feet in this case, then the ball will go over the net.
Let's calculate:
Vx = 42 * cos(16 degrees) = 39.934 ft/s
Vy = 42 * sin(16 degrees) = 11.669 ft/s
t = 2 * 11.669 / 32 = 0.729 s
x = 39.934 * 0.729 = 29.13 ft
y = 2 + (11.669^2) / (2 * 32) = 2 + 48.097 / 64 = 2.752 ft
Since the highest point reached by the ball is lower than the net's height, the ball will not go over the net.