Final answer:
The maximum height of a deer's jump at 3.75 m/s at a 46.3° angle is approximately 0.367 meters, calculated using the initial vertical velocity and the kinematic equations for projectile motion.
Step-by-step explanation:
To find the maximum height of a deer's jump, we need to use the vertical component of the initial velocity and the acceleration due to gravity. Since only the vertical motion affects the maximum height, we can use the formula ½(g)(t^2) = (V_yi)(t), where V_yi is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s2 because it's acting downward), and t is the time it takes for the deer to reach the maximum height (when the vertical velocity is 0).
We first calculate the initial vertical velocity component using V_yi = V*sin(θ), where V is the velocity (3.75 m/s) and θ is the angle of the jump (46.3°). Computing V_yi gives us V_yi = 3.75 m/s * sin(46.3°) = 2.68 m/s (approximately).
Next, we use the kinematic equation to find the time to reach the maximum height, which is when the final vertical velocity becomes zero due to the acceleration of gravity acting against it. Using V_f = V_yi + (g)(t), where V_f is the final velocity (0 m/s at maximum height), we find t = -V_yi / g = -2.68 m/s / -9.8 m/s2 = 0.274 s (approximately).
Now we can find the maximum height with h = (V_yi)(t) + ½(g)(t^2), by plugging in our values for t and V_yi we get h = (2.68 m/s)(0.274 s) + ½(-9.8 m/s2)(0.274 s)2 = 0.367 m or 36.7 cm (approximately). So the maximum height of the deer's jump is about 0.367 meters.