Launch a cannon ball on the ground, at 35 m/sec, at 37 degrees above horizontal.
ID: 1529977 • Letter: L
Question
Launch a cannon ball on the ground, at 35 m/sec, at 37 degrees above horizontal. Find its initial horizontal and vertical components of the velocity Find its horizontal range. Find its maximum height. Find the total time in the air. Find the coordinates (x, y) of the cannon ball, 2 seconds before it strikes the ground Launch a tennis ball at 120 mph, from the ground level. You get horizontal range of 115 m. Fir the initial launch angle. Launch a tennis ball at 100 mph, from the ground level. You get a maximum height of 50 m.Explanation / Answer
1) given
vo = 35 m/s
theta = 37 degrees
Horizontal Range, R = vo^2*sin(2*theta)/g
= 35^2*sin(2*37)/9.8
= 120.1 m
maximum height, Hmax = vo^2*sin^2(theta)/(2*g)
= 35^2*sin^2(37)/(2*9.8)
= 22.6 m
Total time in the air, T = 2*vo*sin(theta)/g
= 2*35*sin(37)/9.8
= 4.30 s
at t = 4.3 - 2 = 2.3 s
x = vox*t
= 35*cos(37)*2.3
= 64.3 m
y = voy*t - (1/2)*g*t^2
= 2.3*35*sin(37) - (1/2)*9.8*2.3^2
= 22.5 m
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