Question

A projectile is fired at a height of 2 meters above the ground with an initial velocity of 100 meteres per second at an angle of 35° with the horizontal. Round each result to the nearest tenths of a unit.

a) Find the vector-valued function describing the motion

b) Find the max height

c) How long was the projectile in the air

d) Find the range

Answer 1

a) R_x = Vcos xt , R_y = Y + Vsinxt + 1/2 ( -gt²)

where R_x = distance travelled when landing horizontally, R_y=distance travelled vertically

Y = Initial height at which body was projected = 2 mtrs.

V = initial velocity = 100 m/s

= Initial angle or projection = 35 degrees

t = time taken before landing

g = gravitational acceleration

b) Max height = R_{y max =} d R_y/ dt = 0+ VSin - gt

now we have to find t.

at max height, V_y=0,

Then, from V_y = VSinxt - gt, putting V_y=0,

t= VSin/g = 100x 0.57/10 = 5.7 sec.

Now, R_y = Y + Vsinxt + 1/2 ( -gt²) = 2 + 100x0.57 - 1/2x10x5.7² = 162.5 mtrs.

c). For R_y = 0,

0 = 2 + 100x0.57xt - 5xt²

=> 5t² - 57t - 2 = 0

=> t = 11.44 sec.

d ) . R_x = Vcos xt = 100 x 0.82 x 11.44 = 938 mtrs.