P4C A ball is thrown at 14.0 m/s from the top of 65.0 m tall building. If the ba

Question

P4C A ball is thrown at 14.0 m/s from the top of 65.0 m tall building. If the ball is thrown at an angle of 20.0 degrees above the horizontal, find: a) The time the ball is in flight. t- b) The horizontal distance from the base of the cliff that the ball strikes the ground. x- We l assume the ground is level in these types of problems unless otherwise specified. If we didn't, we couldn't solve the problem. Time for a little humor: Bumper sticker quote: "Four out of three people have problems with fractions."

Explanation / Answer

1.

Given that

V0 = 14 m/sec at 20 deg with horizontal

V0x = 14*cos 20 deg = 13.16 m/sec

V0y = 14*sin 20 deg = 4.79 m/sec

ay = -g = -9.8 m/sec^2

ax = 0 m/sec^2

Height of building = -65 m

Using 2nd kinematic equation

H = V0y*t + 0.5*ay*t^2

-65 = 4.79*t - 0.5*9.8*t^2

4.9*t^2 - 4.79*t - 65 = 0

Solving quadratic equation

t = [4.79 +/- sqrt (4.79^2 + 4*4.9*65)]/(2*4.9)

taking +ve sign

t = 4.16 sec

Part B.

Range in projectile motion is given by:

R = V0x*t

R = 13.16*4.16

R = 54.74 m

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