A spring loaded cannon hurls a basketball with an initial velocity of 25 m/s. Th

Question

A spring loaded cannon hurls a basketball with an initial velocity of 25 m/s. The cannon is located on the top of a cliff that is 15m above a level field. Suppose the cannon barrel makes an angle of 35 degrees with respect to the horizontal.

What is the maximum height above the field that the basketball will go?

How long will it take to cross the field and hit the ground?

How far across the field will the basketball travel (where will it hit the ground)?

What will the magnitude and direction of the velocity be just prior to hitting the ground?

Explanation / Answer

here,

v0 = 25 m/s

h0 = 15 m

theta = 35 degree

the maximum height above the field that the basketball will go , hmax = ( v0 * sin(theta))^2 /( 2 g)

hmax = ( 25 * sin(35))^2 /( 2 * 9.81) m

hmax = 10.5 m

let the time it take to cross the field and hit the ground be t

h0 = - u * sin(theta) * t+ 0.5 * g * t^2

15 = - 25 * sin(35) * t + 0.5 * 9.81 * t^2

solving for t

t = 3.74 s

the horizontal distance , x = v0 * sin(theta) * t

x = 25 * sin(35) * 3.74 m

x = 53.6 m

the final vertical speed , vy = - u * sin(theta) + g * t

vy = - 25 * sin(35) + 9.81 * 3.74 m/s

vy = 22.35 m/s

horizontal speed , vx = u * cos(theta)

vx = 25 * cos(35) = 20.5 m/s

the magnitude of final speed , |v| = sqrt(vx^2 + vy^2)

|v| = 30.3 m/s

direction , theta = arctan(22.35/30.3) = 36.4 degree

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