angled projectile: for objects launched at an angle, you have to do a little more work to de - termine the…

angled projectile: for objects launched at an angle, you have to do a little more work to de - termine the initial velocity in both the horizontal and vertical directions. for example, if a football is kicked with an initial velocity of 40 m/s at an angle of 30° above the horizontal, you need to break the initial velocity vector up into x - and y - components in the same manner as covered in the components of vectors math review section. then, use the components for your initial velocities in your horizontal and vertical tables. finally, dont forget that symmetry of motion also applies to the parabola of projectile motion. for objects launched and landing at the same height, the launch angle is equal to the landing angle. and if you want an object to travel the maximum possible horizontal distance (or range), launch it at an angle of 45°. 3.52 q: herman the human cannonball is launched from ground level at an angle of 30° above the horizontal with an initial velocity of 26 m/s. how far does herman travel horizontally before reuniting with the ground? 3.55 q: a golf ball is hit at an angle of 45° above the horizontal. what is the acceleration of the golf ball at the highest point in its trajectory? neglect friction. (a) 9.8 m/s² upward (b) 9.8 m/s² downward (c) 6.9 m/s² horizontal (d) 0 m/s².

angled projectile: for objects launched at an angle, you have to do a little more work to de - termine the initial velocity in both the horizontal and vertical directions. for example, if a football is kicked with an initial velocity of 40 m/s at an angle of 30° above the horizontal, you need to break the initial velocity vector up into x - and y - components in the same manner as covered in the components of vectors math review section. then, use the components for your initial velocities in your horizontal and vertical tables. finally, dont forget that symmetry of motion also applies to the parabola of projectile motion. for objects launched and landing at the same height, the launch angle is equal to the landing angle. and if you want an object to travel the maximum possible horizontal distance (or range), launch it at an angle of 45°. 3.52 q: herman the human cannonball is launched from ground level at an angle of 30° above the horizontal with an initial velocity of 26 m/s. how far does herman travel horizontally before reuniting with the ground? 3.55 q: a golf ball is hit at an angle of 45° above the horizontal. what is the acceleration of the golf ball at the highest point in its trajectory? neglect friction. (a) 9.8 m/s² upward (b) 9.8 m/s² downward (c) 6.9 m/s² horizontal (d) 0 m/s².

Answer

3.55 Q

Explanation:

Step1: Analyze horizontal motion

In projectile - motion without air - friction, the horizontal acceleration (a_x = 0) because there is no horizontal force acting on the object (assuming no air - resistance).

Step2: Determine acceleration at highest point

At the highest point of the golf ball's trajectory, the horizontal acceleration remains (a_x=0).

Answer:

D. 0 m/s²

3.52 Q

Explanation:

Step1: Find the horizontal component of the initial velocity

The formula for the horizontal component of the initial velocity is (v_{0x}=v_0\cos\theta), where (v_0 = 26\ m/s) and (\theta = 30^{\circ}). So (v_{0x}=26\cos30^{\circ}=26\times\frac{\sqrt{3}}{2}\approx22.52\ m/s).

Step2: Find the time of flight

The formula for the time of flight (T) of a projectile launched and landing at the same height is (T=\frac{2v_0\sin\theta}{g}), where (g = 9.8\ m/s^{2}), (v_0 = 26\ m/s) and (\theta = 30^{\circ}). So (T=\frac{2\times26\times\sin30^{\circ}}{9.8}=\frac{2\times26\times0.5}{9.8}\approx2.65\ s).

Step3: Calculate the horizontal range

The horizontal range (R) is given by (R = v_{0x}T). Substituting (v_{0x}\approx22.52\ m/s) and (T\approx2.65\ s), we get (R=22.52\times2.65\approx60\ m).

Answer:

Approximately 60 m