Question

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. 3. Approximately how long does it take the projectile to reach the highest point in its trajectory

Answer 1

To calculate the time it takes for a projectile to reach its highest point, use the kinematic equation for vertical motion. With an initial vertical velocity of 40 m/s, and using the acceleration due to gravity, the projectile reaches its highest point in approximately 4.08 seconds.

Step-by-step explanation:

The question pertains to calculating the time it takes for a projectile to reach its highest point in its trajectory, known as the time of ascent. The initial vertical velocity of the projectile is given as 40 m/s. To find the time of ascent, we use the kinematic equation for vertical motion with constant acceleration due to gravity (g = 9.81 m/s2), which is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the velocity at the highest point is 0 m/s (the projectile is momentarily stationary at the highest point), we set v = 0 m/s and solve for t:

0 m/s = 40 m/s + (-9.81 m/s2 × t)

t = 40 m/s / 9.81 m/s2

t ≈ 4.08 seconds

Therefore, it takes approximately 4.08 seconds for the projectile to reach the highest point in its trajectory.

Answer 2

The time is 5.10 sec.

Step-by-step explanation:

Given that,

Component of velocity are 30 m/s and 40 m/s.

We need to calculate the resultant velocity

Using formula of resultant velocity

`v=\sqrt{v_(x)^2+v_(y)^2}`

Put the value into the formula

`v=√((30)^2+(40)^2)`

`v=50\ m/s`[

We need to calculate the time

Using equation of motion

`v = u+gt`

`v = 0+gt`

`t=(v)/(g)`

Put the value into the formula

`t=(50)/(9.8)`

`t=5.10\ sec`

Hence, The time is 5.10 sec.