A cannonball is shot (from ground level) with an initial horizontal velocity of

Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 35 m/s and an initial vertical velocity of 27 m/s.

1)

What is the initial speed of the cannonball?

2)

What is the initial angle ? of the cannonball with respect to the ground?

3)

What is the maximum height the cannonball goes above the ground?

4)

How far from where it was shot will the cannonball land?

5)

What is the speed of the cannonball 1.9 seconds after it was shot?

6)

How high above the ground is the cannonball 1.9 seconds after it is shot?

A cannonball is shot (from ground level) with an initial horizontal velocity of 35 m/s and an initial vertical velocity of 27 m/s. What is the initial speed of the cannonball? What is the initial angle ? of the cannonball with respect to the ground? What is the maximum height the cannonball goes above the ground? How far from where it was shot will the cannonball land? What is the speed of the cannonball 1.9 seconds after it was shot? How high above the ground is the cannonball 1.9 seconds after it is shot?

Explanation / Answer

a)

We know,

velocity = sqrt(horizontal velocity^2 + vertical velocity^2)

initial horizontal velocity, Ux = 35 m/s

initial vertical velocity , Uy = 27 m/s

So, initial velocity = sqrt(Ux^2+Uy^2) = sqrt(35^2+27^2) = 44.2 m/s <-------answer

b)

initial angle = atan(Uy/Ux) = atan(27/35) = 37.6 degrees with respect to ground

c)

maximum height of the cannonball is given by: h = Uy^2/(2*g)

where g = 9.8 m/s2

So, h = 27^2/(2*9.8) = 37.2 m <------answer

d)

using the equation,

v = u +at in vertical direction from the ground to the highest point of its trajectory

where v = final velocity in vertical direction = 0 m/s

u = initial vertical velocity = Uy = 27 m/s

a = acceleration = -9.8 m/s

t = time taken from ground to highest point of its trajectory

So, 0 = 27 - 9.8*t

So, t = 27/9.8 = 2.76 s

So, the time taken by the cannonball to reach the ground again = 2*t = 2*2.76 = 5.52 s

So, the horizontal distance traveled in that time(D) = velocity in horizontal direction(Ux)*time(t)

So, D = 35*5.52 = 193.2 m <-------------answer

e)

using the same equation, v = u + at

in vertical direction, at t = 1.9 s

Vy = Uy + (-9.8)*1.9

So, Vy = 27 - 9.8*1.9 = 8.38 m/s

in horizontal direction, as acceleration = 0 , So, Vx = Ux + 0 = 35 m/s

So, At t = 1.9 s, velocity = sqrt(Vx^2+Vy^2) = sqrt(8.38^2+35^2)

= 35.99 m/s <-------answer

f)

height it goes after t = 1.9 s

Using the equation,

s = ut + 0.5*at^2

where s = height

u = initial vertical velocity = 35 m/s

a = -9.8

So,at t = 1.9 s

s = 35*1.9 + 0.5*(-9.8)*1.9^2 = 48.8 m <------------answer

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