A stone is thrown upward from the top of building at an angle of 30 degree to th

Question

A stone is thrown upward from the top of building at an angle of 30 degree to the horizontal and with an initial speed of 20 m/s. The point of release is H, 48 m above the ground? Let g = 10.0 m/s2. What is the maximum height from the ground? How long does it take for the stone to hit the ground? After 2 second, obtain the x coordinate. After 2 second, obtain the y coordinate Find the stone's speed at impact Find the horizontal range R of the stone. Find the velocity of the stone at the place 24 m above the ground. Find the time at that place.

Explanation / Answer

u = 20; g = 10

h0 = 48

a) hmax = u*u*sin(30)*sin(30)/(2*g) + h0

hmax = 48 +5

hmax = 53 m/x

b) -48 = u*sin(30)*t - 0.5*g*t*t

t = 4.256 m/s

c) x = u*cos(30)*t1

t1 = 2

x = 34.64 m

d) y = h0 + u*sin(30)*t1 - 0.5*g*t1*t1

y = 48 m

e) impact takes place at t = 4.256 seconds (using part (b))

vx = u*cos(30)

vy = u*sin(30) - g*t

v^2 = vx^2 + vy^2

v = 36.88 m/s

f) R = u*cos(30)*t

R = 73.71 m

g) -24 = u*sin(30)*t3 - 0.5*g*t3*t3

vx = u*cos(30)

vy = u*sin(30) - g*t3

v^2 = vx^2 + vy^2

v = 29.66 m/s

h) -24 = u*sin(30)*t3 - 0.5*g*t3*t3

t3 = 3.408 seconds

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