Question

A projectile is fired with an initial speed of 49.6 m/s at an angle of 45.2 ∘ above the horizontal on a long flat firing range.

Determine the speed of the projectile 1.50 s after firing.
Determine the direction of the motion of the projectile 1.50 s after firing.

Answer 1

The projectile's speed and direction 1.50 s after firing are determined by calculating the horizontal and vertical components of velocity,

where the horizontal component remains constant and the vertical component is influenced by gravity. The speed is found using the Pythagorean theorem and the direction with the arctangent function.

Step-by-step explanation:

To determine the speed and direction of the projectile 1.50 s after firing, we must consider the components of the initial velocity separately.

The initial velocity can be broken down into horizontal (vx0) and vertical (vy0) components using trigonometry, which are vx0 = v0 × cos(θ) and vy0 = v0 × sin(θ) respectively, where v0 = 49.6 m/s and θ = 45.2°.

The horizontal velocity remains constant because there is no horizontal acceleration, so vx = vx0. The vertical velocity, however, changes due to the acceleration due to gravity (g = 9.81 m/s2), calculated as vy = vy0 - g × t, where t = 1.50 s. To find the speed at 1.50 s, we combine these velocities using the Pythagorean theorem: speed = √(vx2 + vy2).

For the direction, we can use the arctangent function: direction = arctan(vy/vx), which gives the angle the velocity vector makes with the horizontal.

Since the horizontal velocity is constant and the vertical velocity is only affected by gravity, we expect the direction will have decreased from the initial 45.2° angle due to the downward acceleration.