A projectile is fired from ground level with an initial speed v0 = 72.3 m/s at a

Question

A projectile is fired from ground level with an initial speed v0 = 72.3 m/s at an angle theta = 69.2 degree above the horizontal. (a) Calculate the time necessary for the projectile to reach its maximum height. (b) Calculate the maximum height h reached by the projectile. (c) Calculate the horizontal and vertical components of the velocity vector at the maximum height. (d) Calculate horizontal distance D travelled by projectile, if the landing point is 20 m below the starting ground level. (e) Calculate the magnitude of the velocity at that landing point, and the angle it makes with the horizontal.

Explanation / Answer

a) max height when vy = 0

vy = v0y + a t

0 = 72.3*sin(69.2 degrees) - 9.81*t

t=6.89 s

b)

vy^2 = v0y^2 + 2 a x

0 = (72.3*sin(69.2 degrees))^2 - 2*9.81*y

y=232.8 m

c)at max height vy = 0

vx doesnt change so vx = 72.3*cos(69.2 degrees)= 25.67 m/s

d) y direction

y = y0 + v0y t + 1/2 a t^2

-20 = 0 + 72.3*sin(69.2 degrees)*t - 0.5*9.81*t^2

t=14.07 s

now x direction
x = v0x t = 25.67*14.07= 361.2 m

e) find vy when landing

vy = v0y + a t= 72.3*sin(69.2 degrees) - 9.81*14.07= -70.4 m/s

mag = sqrt(70.4^2 + 25.67^2)= 74.9 m/s

angle = arctan(vy/vx) = arctan(70.4/25.67)= 70.0 degrees below horizontal

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