Question

Suppose the ball is thrown from the height h = 43.0 m at an angle of 34.0°
below the horizontal. If it strikes the ground 41.8 m away, find the following. (Hint: For part (a), use the
equation for the x displacement to eliminate v0t from the equation for the y displacement.)

(a) the time of flight

(b) the initial speed
m/s
(c) the speed and angle of the velocity vector with respect to the horizontal at impact

speed m/s

angle ° below the horizontal

Answer 1

y = -43m; x = 41.8 m ; @ = -34 deg ;

using equation of path

y = xtan@ - 4.9x^2 / (Vcos@)^2

-43 = 41.8tan(-34) - 4.9 x 41.8^2 / (Vcos34)^2

V = 29 m/s initial speed

V^2 = u^2 + 2as

= 29^2 + 2x9.8x43

V = 41 m/s final speed

@ = cos^-1(29cos34/41) = 54.1 deg

V = u + at

-41sin54.1 = -29sin34 - 9.8 t

t = 5.04 sec