A home run is hit such a way that the baseball just clears a wall 30 m high located 117 m from home plate. The ball is hit at an angle of 40 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 2 m above the ground. The acceleration of gravity is 9.8 m/s^2.
What is the initial speed of the ball?
Answer in units of m/s
You need to solve two simultaneous equations for the time of flight (T) and initial velocity, Vo.
Vo cos40 T = 117 meters
or
Vo*T = 152.7 m,
and
2 + Vo sin40 T- (g/2) T^2 = 30 meters
or
0.6428 Vo*T - 4.9 T^2 = 28
Solve for T first, then Vo
98.17 -4.9 T^2 = 28
T = 3.99 s
etc.
Vi=60m/s
To find the initial speed of the ball, we can use the kinematic equation that relates the range (horizontal distance traveled) of a projectile to its initial speed and launch angle. The range, in this case, is the horizontal distance from home plate to the wall, which is 117 m.
The range equation is given by:
R = (v^2 * sin(2θ)) / g
Where:
R is the range
v is the initial speed of the ball
θ is the launch angle
g is the acceleration due to gravity
In this case, the range (R) is equal to 117 m, the launch angle (θ) is 40 degrees, and the acceleration due to gravity (g) is 9.8 m/s^2.
Let's substitute these values into the equation and solve for v.
117 m = (v^2 * sin(2*40°)) / 9.8 m/s^2
To simplify further, we can use the sine double-angle identity: sin(2θ) = 2 * sin(θ) * cos(θ)
117 m = (v^2 * 2 * sin(40°) * cos(40°)) / 9.8 m/s^2
Now we can rearrange the equation to solve for v:
v^2 = (117 m * 9.8 m/s^2) / (2 * sin(40°) * cos(40°))
v^2 = (1146.6 m^2/s^2) / (2 * 0.6428 * 0.7660)
v^2 = (1146.6 m^2/s^2) / 0.9845
v^2 ≈ 1165.14
Taking the square root of both sides to solve for v:
v ≈ √1165.14
v ≈ 34.12 m/s
Therefore, the initial speed of the ball is approximately 34.12 m/s.