A projectile is fired from the ground (you can assume the initial height is the

Question

A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no obstacles in its way. It is fired at an angle 35° with respect to the horizontal and with an initial speed of 38 m/s. Air resistance is negligible in this situation.
a)What is the maximum height that the projectile reaches?

b)What is the speed of the projectile at its maximum height?

The same projectile was then fired in the same way toward a wall that is a horizontal distance 84.3 m from where the projectile was fired.

c)What was the height of the projectile when it hit the wall?

d)What was the velocity of the projectile when it hit the wall?
Call down the positive y direction, and toward the wall the positive x direction.
i +   j;

e)What was the speed of the projectile when it hit the wall?

f)If the wall were slightly farther from where the projectile was launched (the projectile still hits it), how would that effect the following? ( Increases Decreases Stays the same)

The maximum height of the projectile.
The magnitude of the y component of the velocity of the projectile when it hits the wall.
The distance from the ground to where the projectile hit the wall.
The magnitude of the x component of the velocity of the projectile when it hits the wall.
The speed of the projectile when it hits the wall.

Explanation / Answer

a)

Given that

The initial speed of the projectile fired from the gun (u) =38m/s

The projectile fired with the gun make an angle theta with the horizontal (theta) =35degrees

Acceleration due to gravity (g) =9.81m/s2

The maximum height reached by the projectile is

H =u2sin2theta/2g =(38)2*sin2(35)/2*9.81=24.213m

b)

The speed of the projectile at the maximum height is given by Velocity in projectile motion has two components : horizontal and vertical. At the highest point of the motion, the vertical velocity would become zero and horintal velocity would be u(h)= ucos(x), x is the angle of the projectile and u(h)= horizontal velocity and u = the velocity of the projectile. At the highest point the vertical velocity is 0.

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