Question

A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of the trajectory? (b) What is its time of flight?

Answer 1

Part a)

`v_f = v_x = 32.77 m/s`

Part b)

T = 4.68 s

Step-by-step explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

`v_x = 40 cos35 = 32.77 m/s`

`v_y = 40 sin35 = 22.94 m/s`

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

`v_f = 32.77 m/s`

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

`\Delta y = 0`

`0 = v_y t - (1)/(2)gt^2`

`T = (2v_y)/(g)`

`T = (2(22.94))/(9.81) = 4.68 s`