a ball is thrown at an angel of 40 degee above tghe horizontal.the ball clears the fence whis is 85 metre away.what is the inital velosity of the ball if we assume the ball is thrown at the same elevation of the fence
u = S cos 40 = .766 S
Vi = S sin 40 = .643 S
h = Vi t - 4.9 t^2 = 0 at start and at fence
t(4.9 t -Vi) = 0
t = Vi/4.9 = .131 S seconds to fence
d = u t = .766 S (.131 S) = .1 S^2
so
.1 S^2 = 85
S^2 = 850
S = 29.2 m/s
To find the initial velocity of the ball, we can use the kinematic equation for projectile motion:
d = (v0^2 * sin(2θ)) / g
where:
d is the horizontal distance traveled by the ball (85 meters),
v0 is the initial velocity of the ball,
θ is the launch angle (40 degrees), and
g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation to solve for v0, we get:
v0 = sqrt((d * g) / sin(2θ))
Now, substituting in the given values:
v0 = sqrt((85 * 9.8) / sin(2 * 40))
Using a scientific calculator, calculate the value of sin(2 * 40) and then substitute it into the equation:
v0 = sqrt((85 * 9.8) / 0.6428)
v0 = sqrt(835.3 / 0.6428)
v0 = sqrt(1298.9)
v0 ≈ 36.03
Therefore, the initial velocity of the ball is approximately 36.03 meters per second.