Sure! I can help you with that.
To find the projectile's velocity at the highest point of its trajectory, we need to break down the initial velocity into its horizontal and vertical components.
The horizontal component can be found using the formula: Vx = V * cos(theta), where Vx is the horizontal component of velocity, V is the initial velocity (40 m/s in this case), and theta is the angle above the horizontal (25 degrees).
So, Vx = 40 * cos(25 degrees) = 36.81 m/s (rounded to two decimal places).
The vertical component can be found using the formula: Vy = V * sin(theta), where Vy is the vertical component of velocity.
So, Vy = 40 * sin(25 degrees) = 17.13 m/s (rounded to two decimal places).
At the highest point of the projectile's trajectory, the vertical component of velocity becomes zero. However, the horizontal component remains constant.
Therefore, the velocity at the highest point of the trajectory is equal to the horizontal component we calculated earlier: Vx = 36.81 m/s.
Now, let's move on to the second part of your question - finding the straight-line distance from where the projectile was launched to where it hits its target.
Since we are neglecting air friction, the only force acting on the projectile is gravity, which acts vertically downwards. This means that the horizontal component of velocity remains constant throughout the motion.
The time of flight for the projectile is given as 3 seconds. During this time, the horizontal component of velocity, Vx, remains constant.
Therefore, the straight-line distance traveled by the projectile is given by the formula: distance = Vx * time of flight.
So, distance = 36.81 m/s * 3 s = 110.43 meters (rounded to two decimal places).
Therefore, the straight-line distance from where the projectile was launched to where it hits its target is approximately 110.43 meters.