Answer:
To calculate the full distance traveled by an object thrown at a velocity of 35 m/s at an angle of 45 degrees, we need to consider the horizontal and vertical components of the motion separately.
The horizontal component of the motion remains constant throughout the trajectory and is given by:
Horizontal distance = (Initial velocity) * (Time of flight) * cos(angle)
In this case, the initial velocity is 35 m/s, the angle is 45 degrees, and we need to find the time of flight.
The time of flight can be calculated using the vertical component of the motion. The vertical motion can be described using the equation:
Vertical displacement = (Initial velocity * sin(angle))^2 / (2 * acceleration)
Where the initial velocity is 35 m/s, the angle is 45 degrees, and the acceleration is the acceleration due to gravity, approximately 9.8 m/s^2.
The vertical displacement is zero at the highest point of the trajectory since the object comes back down to the same height it was launched from. So we can solve the equation for the time of flight.
Using these calculations, we can find the horizontal distance traveled by the object.
Let's calculate step by step:
Step 1: Calculate the time of flight
Vertical displacement = 0 (at the highest point)
0 = (35 * sin(45))^2 / (2 * 9.8)
0 = (24.75^2) / 19.6
0 = 616.0125 / 19.6
0 = 31.43
Step 2: Calculate the time of flight
Vertical displacement = (Initial velocity * sin(angle)) * time - (1/2) * acceleration * time^2
0 = (35 * sin(45)) * time - (1/2) * 9.8 * time^2
0 = 24.75 * time - 4.9 * time^2
4.9 * time^2 - 24.75 * time = 0
time * (4.9 * time - 24.75) = 0
time = 0 (initial point) or 24.75 / 4.9
time = 5.05 seconds
Step 3: Calculate the horizontal distance
Horizontal distance = (Initial velocity) * (Time of flight) * cos(angle)
Horizontal distance = 35 * 5.05 * cos(45)
Horizontal distance = 35 * 5.05 * (sqrt(2)/2)
Horizontal distance = 88.96 meters
Therefore, when an object is thrown at 35 m/s at an angle of 45 degrees, the full distance traveled is approximately 88.96 meters.